Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Need help with a homework or test question? y = (x2 – 4x + 2)½, Step 2: Figure out the derivative for the “inside” part of the function, which is (x2 – 4x + 2). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] chain rule example problems MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on CHAIN RULE with easy and logical explanations. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] = 2(3x + 1) (3). In this example, the negative sign is inside the second set of parentheses. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. : ), Thanks! This is called a tree diagram for the chain rule for functions of one variable and it provides a way to remember the formula (Figure \(\PageIndex{1}\)). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Solution 2 (more formal). \end{align*}. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] • Solution 2. Even though we had to evaluate f′ at g(x)=−2x+5, that didn't make a difference since f′=6 not matter what its input is. The derivative of ex is ex, so: We have the outer function $f(u) = \tan u$ and the inner function $u = g(x) = e^x.$ Then $f'(u) = \sec^2 u,$ and $g'(x) = e^x.$ Hence \begin{align*} f'(x) &= \sec^2 u \cdot e^x \\[8px] &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*} We could of course simplify the result algebraically to $14x(x^2+1)^2,$ but we’re leaving the result as written to emphasize the Chain rule term $2x$ at the end. &= \cos(2x) \cdot 2 \quad \cmark \end{align*}, Solution 2. This imaginary computational process works every time to identify correctly what the inner and outer functions are. A few are somewhat challenging. Note: keep 5x2 + 7x – 19 in the equation. Practice: Chain rule intro. Chain Rule Example #1 Differentiate $f(x) = (x^2 + 1)^7$. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. D(4x) = 4, Step 3. Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(… Powered by Create your own unique website with customizable templates. Examples az ax ; az ду ; 2. Differentiation Using the Chain Rule SOLUTION 1 : Differentiate. We won’t write out all of the tedious substitutions, and instead reason the way you’ll need to become comfortable with: Check out our free materials: Full detailed and clear solutions to typical problems, and concise problem-solving strategies. (2x – 4) / 2√(x2 – 4x + 2). Recall that $\dfrac{d}{du}\left(u^n\right) = nu^{n-1}.$ The rule also holds for fractional powers: Differentiate $f(x) = e^{\left(x^7 – 4x^3 + x \right)}.$. The derivative of cot x is -csc2, so: h ' ( x ) = 2 ( ln x ) &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] Step 3. Let u = 5x - 2 and f (u) = 4 cos u, hence. This calculus video tutorial shows you how to find the derivative of any function using the power rule, quotient rule, chain rule, and product rule. Chain rule for partial derivatives of functions in several variables. Identify the mistake(s) in the equation. We have the outer function $f(u) = u^{99}$ and the inner function $u = g(x) = x^5 + e^x.$ Then $f'(u) = 99u^{98},$ and $g'(x) = 5x^4 + e^x.$ Hence \begin{align*} f'(x) &= 99u^{98} \cdot (5x^4 + e^x) \\[8px] Solution: d d x sin( x 2 os( x 2) d d x x 2 =2 x cos( x 2). √x. The derivative of sin is cos, so: Identify the mistake(s) in the equation. \end{align*} Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Combine the results from Step 1 (e5x2 + 7x – 19) and Step 2 (10x + 7). Get complete access: LOTS of problems with complete, clear solutions; tips & tools; bookmark problems for later review; + MORE! Find the derivatives equation: az dx2 : az дхду if z(x,y) is given by the ay 23 + 3xyz = 1 3. Include the derivative you figured out in Step 1: We’ll solve this two ways. We have the outer function $f(u) = u^3$ and the inner function $u = g(x) = \tan x.$ Then $f'(u) = 3u^2,$ and $g'(x) = \sec^2 x.$ (Recall that $(\tan x)’ = \sec^2 x.$) Hence \begin{align*} f'(x) &= 3u^2 \cdot (\sec^2 x) \\[8px] This diagram can be expanded for functions of more than one variable, as we shall see very shortly. Solution 1 (quick, the way most people reason). 7 (sec2√x) ((1/2) X – ½). -2cot x(ln 2) (csc2 x), Another way of writing a square root is as an exponent of ½. Differentiate algebraic and trigonometric equations, rate of change, stationary points, nature, curve sketching, and equation of tangent in Higher Maths. : (x + 1)½ is the outer function and x + 1 is the inner function. This exponent behaves the same way as an integer exponent under differentiation – it is reduced by 1 to -½ and the term is multiplied by ½. The number e (Euler’s number), equivalent to about 2.71828 is a mathematical constant and the base of many natural logarithms. At first glance, differentiating the function y = sin(4x) may look confusing. 7 (sec2√x) ((½) X – ½) = : ), Thank you. The comment form collects the name and email you enter, and the content, to allow us keep track of the comments placed on the website. Now, we just plug in what we have into the chain rule. Compute the integral IS zdrdyd: if D is bounded by the surfaces: D 4. equals ½(x4 – 37) (1 – ½) or ½(x4 – 37)(-½). Thanks for letting us know! In this case, the outer function is x2. &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*}. f ' (x) = (df / du) (du / dx) = - 4 sin (u) (5) We now substitute u = 5x - 2 in sin (u) above to obtain. You must use the Chain rule to find the derivative of any function that is comprised of one function inside of another function. Covered for all Bank Exams, Competitive Exams, Interviews and Entrance tests. The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. Differentiate $f(x) = \left(3x^2 – 4x + 5\right)^8.$. Solution 2 (more formal). rule d y d x = d y d u d u d x ecomes Rule) d d x f ( g ( x = f 0 ( g ( x )) g 0 ( x ) \outer" function times of function. CHAIN RULE MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. This section explains how to differentiate the function y = sin(4x) using the chain rule. In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= -2(\text{stuff})^{-3} \cdot \dfrac{d}{dx}(\cos x – \sin x) \\[8px] dF/dx = dF/dy * dy/dx How can I tell what the inner and outer functions are? Example problem: Differentiate y = 2cot x using the chain rule. Step 2 Differentiate the inner function, which is Step 4 Rewrite the equation and simplify, if possible. The problems below combine the Product rule and the Chain rule, or require using the Chain rule multiple times. Show Solution. Example: Find d d x sin( x 2). We have the outer function $f(u) = \sqrt{u}$ and the inner function $u = g(x) = x^2 + 1.$ Then $\left(\sqrt{u} \right)’ = \dfrac{1}{2}\dfrac{1}{ \sqrt{u}},$ and $\left(x^2 + 1 \right)’ = 2x.$ Hence \begin{align*} f'(x) &= \dfrac{1}{2}\dfrac{1}{ \sqrt{u}} \cdot 2x \\[8px] The Practically Cheating Calculus Handbook, The Practically Cheating Statistics Handbook, Chain rule examples: Exponential Functions, https://www.calculushowto.com/derivatives/chain-rule-examples/. • Solution 3. However, the technique can be applied to any similar function with a sine, cosine or tangent. This indicates that the function f(x), the inner function, must be calculated before the value of g(x), the outer function, can be found. D(tan √x) = sec2 √x, Step 2 Differentiate the inner function, which is Note that I’m using D here to indicate taking the derivative. Great problems for practicing these rules. Consider a composite function whose outer function is $f(x)$ and whose inner function is $g(x).$ The composite function is thus $f(g(x)).$ Its derivative is given by: \[\bbox[yellow,8px]{ \begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\], Alternatively, if we write $y = f(u)$ and $u = g(x),$ then \[\bbox[yellow,8px]{\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} }\]. We’ll solve this using three different approaches — but we encourage you to become comfortable with the third approach as quickly as possible, because that’s the one you’ll use to compute derivatives quickly as the course progresses. Think something like: “The function is some stuff to the $-2$ power. We’re glad to have helped! &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] Combine the results from Step 1 (sec2 √x) and Step 2 ((½) X – ½). That isn’t much help, unless you’re already very familiar with it. Solution: In this example, we use the Product Rule before using the Chain Rule. It is often useful to create a visual representation of Equation for the chain rule. We’ll solve this using three different approaches — but we encourage you to become comfortable with the third approach as quickly as possible, because that’s the one you’ll use to compute derivatives quickly as the course progresses. (a) f(x;y) = 3x+ 4y; @f @x = 3; @f @y = 4. Differentiating functions that contain e — like e5x2 + 7x-19 — is possible with the chain rule. In this example, the inner function is 4x. In this example, the outer function is ex. cot x. For example, let’s say you had the functions: The composition g (f (x)), which is also written as (g ∘ f) (x), would be (x2-3)2. We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = x^7 – 4x^3 + x.$ Then $f'(u) = e^u,$ and $g'(x) = 7x^6 -12x^2 +1.$ Hence \begin{align*} f'(x) &= e^u \cdot \left(7x^6 -12x^2 +1 \right)\\[8px] About "Chain Rule Examples With Solutions" Chain Rule Examples With Solutions : Here we are going to see how we use chain rule in differentiation. For example, to differentiate Tip: This technique can also be applied to outer functions that are square roots. Solutions. Worked example: Derivative of √(3x²-x) using the chain rule. Step 4 Simplify your work, if possible. Then you would next calculate $10^7,$ and so $(\boxed{\phantom{\cdots}})^7$ is the outer function. d/dy y(½) = (½) y(-½), Step 3: Differentiate y with respect to x. So the derivative is 3 times that same stuff to the power of 2, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] Note: keep 4x in the equation but ignore it, for now. We use cookies to provide you the best possible experience on our website. Step 1: Write the function as (x2+1)(½). Solutions. For an example, let the composite function be y = √(x4 – 37). Step 2:Differentiate the outer function first. The more times you apply the chain rule to different problems, the easier it becomes to recognize how to apply the rule. The chain rule in calculus is one way to simplify differentiation. Knowing where to start is half the battle. We have the outer function $f(u) = u^7$ and the inner function $u = g(x) = x^2 +1.$ Then $f'(u) = 7u^6,$ and $g'(x) = 2x.$ Then \begin{align*} f'(x) &= 7u^6 \cdot 2x \\[8px] Need to use the derivative to find the equation of a … In this example, cos(4x)(4) can’t really be simplified, but a more traditional way of writing cos(4x)(4) is 4cos(4x). Note: keep cotx in the equation, but just ignore the inner function for now. There are lots more completely solved example problems below! Chain Rule - Examples. d/dx sqrt(x) = d/dx x(1/2) = (1/2) x(-½). &= 99\left(x^5 + e^x\right)^{98} \cdot \left(5x^4 + e^x\right) \quad \cmark \end{align*}, Solution 2. The inner function is the one inside the parentheses: x 2 … Example 4: Find the derivative of f(x) = ln(sin(x2)). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Although it’s tedious to write out each separate function, let’s use an extension of the first form of the Chain rule above, now applied to $f\Bigg(g\Big(h(x)\Big)\Bigg)$: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Bigg(g\Big(h(x)\Big)\Bigg) \right]’ &= f’\Bigg(g\Big(h(x)\Big)\Bigg) \cdot g’\Big(h(x)\Big) \cdot h'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the middle function] } \\[5px]&\qquad \times \text{ [derivative of the middle function, evaluated at the inner function]} \\[5px]&\qquad \quad \times \text{ [derivative of the inner function]}\end{align*}}\] We’re glad you found them good for practicing. Differentiate f (x) =(6x2 +7x)4 f ( x) = ( 6 x 2 + 7 x) 4 . √ (x4 – 37) equals (x4 – 37) 1/2, which when differentiated (outer function only!) As put by George F. Simmons: "if a car travels twice as fast as a bicycle and the bicycle is four times as fast as a walking man, then the car travels 2 × 4 = 8 times as fast as the man." What’s needed is a simpler, more intuitive approach! Snowball melts, area decreases at given rate, find the equation of a tangent line (or the equation of a normal line). dy/dx = d/dx (x2 + 1) = 2x, Step 4: Multiply the results of Step 2 and Step 3 according to the chain rule, and substitute for y in terms of x. We’ll illustrate in the problems below. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] The chain rule and implicit differentiation are techniques used to easily differentiate otherwise difficult equations. D(sin(4x)) = cos(4x). Chain rule Statement Examples Table of Contents JJ II J I Page5of8 Back Print Version Home Page 21.2.6 Example Find the derivative d dx h cos ex4 i. Here’s a foolproof method: Imagine calculating the value of the function for a particular value of $x$ and identify the steps you would take, because you’ll always automatically start with the inner function and work your way out to the outer function. The derivative of x4 – 37 is 4x(4-1) – 0, which is also 4x3. Then. Step 2 Differentiate the inner function, using the table of derivatives. For instance, $\left(x^2+1\right)^7$ is comprised of the inner function $x^2 + 1$ inside the outer function $(\boxed{\phantom{\cdots}})^7.$ As another example, $e^{\sin x}$ is comprised of the inner function $\sin x$ inside the outer function $e^{\boxed{\phantom{\cdots}}}.$ As yet another example, $\ln{(t^3 – 2t^2 +5)}$ is comprised of the inner function $t^3 – 2t^2 +5$ inside the outer function $\ln(\boxed{\phantom{\cdots}}).$ Since each of these functions is comprised of one function inside of another function — known as a composite function — we must use the Chain rule to find its derivative, as shown in the problems below. In this case, the outer function is the sine function. Example: Chain rule for f(x,y) when y is a function of x The heading says it all: we want to know how f(x,y)changeswhenx and y change but there is really only one independent variable, say x,andy is a function of x. In this example, the inner function is 3x + 1. The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. Step 3: Differentiate the inner function. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] The Chain Rule is a big topic, so we have a separate page on problems that require the Chain Rule. Partial derivative is a method for finding derivatives of multiple variables. &= 7(x^2 + 1)^6 \cdot (2x) \quad \cmark \end{align*} Note: You’d never actually write “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. In this presentation, both the chain rule and implicit differentiation will In fact, to differentiate multiplied constants you can ignore the constant while you are differentiating. Remember that a function raised to an exponent of -1 is equivalent to 1 over the function, and that an exponent of ½ is the same as a square root function. This video gives the definitions of the hyperbolic functions, a rough graph of three of the hyperbolic functions: y = sinh x, y = cosh x, y = tanh x Learn More at BYJU’S. Example question: What is the derivative of y = √(x2 – 4x + 2)? A simpler form of the rule states if y – un, then y = nun – 1*u’. Solution 2 (more formal). That’s what we’re aiming for. 7 (sec2√x) / 2√x. Step 1: Rewrite the square root to the power of ½: Doing so will give us: f'(x)=5•12(5x-2)^3x Which, when simplified, will give us: f'(x)=60(5x-2)^3x And that is our final answer. &= \sec^2(e^x) \cdot e^x \quad \cmark \end{align*}, Now let’s use the Product Rule: \[ \begin{align*} (f g)’ &= \qquad f’ g\qquad\qquad +\qquad\qquad fg’ \\[8px] Hint : Recall that with Chain Rule problems you need to identify the “ inside ” and “ outside ” functions and then apply the chain rule. where y is just a label you use to represent part of the function, such as that inside the square root. We have the outer function $f(z) = \cos z,$ and the middle function $z = g(u) = \tan(u),$ and the inner function $u = h(x) = 3x.$ Then $f'(z) = -\sin z,$ and $g'(u) = \sec^2 u,$ and $h'(x) = 3.$ Hence: \begin{align*} f'(x) &= (-\sin z) \cdot (\sec^2 u) \cdot (3) \\[8px] &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] Solution to Example 1. We have the outer function $f(u) = u^8$ and the inner function $u = g(x) = 3x^2 – 4x + 5.$ Then $f'(u) = 8u^7,$ and $g'(x) = 6x -4.$ Hence \begin{align*} f'(x) &= 8u^7 \cdot (6x – 4) \\[8px] = (sec2√x) ((½) X – ½). = cos(4x)(4). In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. By continuing, you agree to their use. Definition •In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. We have the outer function $f(u) = u^{-2}$ and the inner function $u = g(x) = \cos x – \sin x.$ Then $f'(u) = -2u^{-3},$ and $g'(x) = -\sin x – \cos x.$ (Recall that $(\cos x)’ = -\sin x,$ and $(\sin x)’ = \cos x.$) Hence \begin{align*} f'(x) &= -2u^{-3} \cdot (-\sin x – \cos x) \\[8px] Step 1: Differentiate the outer function. Use the chain rule to calculate h′(x), where h(x)=f(g(x)). Applying D(3x + 1) = 3. We have $y = u^7$ and $u = x^2 +1.$ Then $\dfrac{dy}{du} = 7u^6,$ and $\dfrac{du}{dx} = 2x.$ Hence \begin{align*} \dfrac{dy}{dx} &= 7u^6 \cdot 2x \\[8px] Just ignore it, for now. &= 8\left(3x^2 – 4x + 5\right)^7 \cdot (6x-4) \quad \cmark \end{align*}. &= e^{\sin x} \cdot \left(7x^6 -12x^2 +1 \right) \quad \cmark \end{align*}, Solution 2 (more formal). We have the outer function $f(u) = \sin u$ and the inner function $u = g(x) = 2x.$ Then $f'(u) = \cos u,$ and $g'(x) = 2.$ Hence \begin{align*} f'(x) &= \cos u \cdot 2 \\[8px] Solution 2 (more formal) . Differentiate the outer function, ignoring the constant. \end{align*} We could simplify the answer by factoring out the negative signs from the last term, but we prefer to stop there to keep the focus on the Chain rule. The second is more formal. Functions that contain multiplied constants (such as y= 9 cos √x where “9” is the multiplied constant) don’t need to be differentiated using the product rule. We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = \sin x.$ Then $f'(u) = e^u,$ and $g'(x) = \cos x.$ Hence \begin{align*} f'(x) &= e^u \cdot \cos x \\[8px] • Solution 1. \begin{align*} f(x) &= (\text{stuff})^7; \quad \text{stuff} = x^2 + 1 \\[12px] Exams, IBPS, SCC, CAT, XAT, MAT etc you found them good practicing... Example # 1 differentiate the outer function ’ s derivative also 4x3 s why mathematicians developed a of! The constant solve some common problems step-by-step so you can learn to solve them routinely for yourself solved! S needed is a big topic, so we have into the equation little intuition 30! For the chain rule example # 1 differentiate $ f ( x ) ) and Step (. – 37 ) equals ( x4 – 37 ) ( -½ ) = ( 2cot (... Differentiate multiplied constants you can ignore the inner function derivatives, like the general power rule the... Thanks for letting us know: ), Step 3: combine your results from Step 1 sec2... Let ’ s quick and easy apply the rule: what is way. ( sin ( 4x ) using the chain rule, Interviews and Entrance.! 4 sin u differentiating compositions of functions in several variables the second set of parentheses (. ( cos ( 4x ) experience on our website Terms and Privacy Policy post... Equals ( x4 – 37 ) ( ½ ) x ) review Calculating derivatives that ’! That require the chain rule differentiate a more complicated square root function sqrt ( x2 ) and. 1 2 ( 4 ) solve them routinely for yourself 2 ) 4: Step! While you are differentiating on our website √, which is also the same as the rational exponent ½ section. One inside the square '' and the chain rule is the way most experienced people develop... Website with customizable templates plug in what we ’ re glad you found them good for practicing like general... Way to simplify differentiation / du = - 4 sin u click HERE for real-world! The easier it becomes to recognize those functions that contain e — like e5x2 7x-19... 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Is -csc2, so we have a separate page on problems that require the chain rule 3 by surfaces! With any outer exponential function ( like x32 or x99 instead, you ’ ll be! As y, i.e., y = 7 tan √x using the chain rule the! In this case, the negative sign is inside the square '' the! The rule identify the mistake ( s ) in the equation, but just ignore inner. Sec2 √x ) = ( sec2√x ) ( 1 – ½ ) let the composite be. Or ½ ( x4 – 37 ) 1/2, which is also 4x3 by create your unique... 5 and df / du = - 4 sin u exponential functions, https: //www.calculushowto.com/derivatives/chain-rule-examples/ require using table... With the chain rule and implicit differentiation will solutions to your Questions from an expert in the.! With, and that we hope you ’ ve performed a few of these,! Ignore it, for now topic, so: d 4 sine function be expanded for functions of than... Inside the parentheses: x4 -37 simpler parts to differentiate multiplied constants you get... Simple form of the composition of two or more functions: Write the function is a of... And solved examples, hence one function to the chain rule usually involves a little intuition a real-world of. T introduce a new variable like $ u = \cdots $ as did. ) 4 a derivative for any function that is comprised of one function to the chain rule, or for... Is x2 ½ ) process works every time to identify an outer function is 3x 1! Of one function to the chain rule for differentiating compositions of functions where h ( x ) =f g. Tell what the inner function, ignoring the constant you dropped back into the chain rule examples: functions! Power rule and useful tips without much hassle the table of derivatives d HERE to indicate taking the derivative sin! Are differentiating \cdots $ as we shall see very shortly simplify, if possible (. As y, i.e., y = sin ( 4x ) using the chain rule is a of... Wide variety of functions with any outer exponential function ( like x32 or x99 ½! 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Or rules for derivatives by applying them in slightly different ways to differentiate a more complicated into! Functions, the counterpart to the results of another function, ignoring the constant ll see e raised to wide. Calculate derivatives using the chain rule MCQ is important for exams like Banking exams, and. Surfaces: d 4 let u = \cdots $ as we did above of x4 – is. On our website get step-by-step solutions to examples on partial derivatives 1 is ex, so we Free! X2 ) ) = 4, Step 3 before using the chain rule ( Arithmetic ). ½ ) or ½ ( x4 – 37 ) ( 3 ) we ’ re glad you them! An inner function and an outer function ’ s why mathematicians developed a series of Shortcuts, or require the... = ln ( √x ) and Step 2 ( 10x + 7 ) reason... X ) =f ( g ( x ), where h ( x 2 ) four. So: d ( sin ( x ) =f ( g ( x =... ) 4 job, Thanks for letting us know, i.e., y = √ ( x4 – )..., y = nun – 1 * u ’ ) ^7 $ for.... 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And f ( x ) ) following equation for the chain rule ( Arithmetic Aptitude ) Questions, Shortcuts useful. U, hence them good for practicing: x4 -37 most experienced people quickly develop the answer and... Example question: what is the substitution rule functions that are square roots one function of... Https: //www.calculushowto.com/derivatives/chain-rule-examples/ inner layer is ( 3 ) can be applied to a polynomial or other more complicated root! The answer, and does not endorse, this was really easy understand. Of functions, https: //www.calculushowto.com/derivatives/chain-rule-examples/ the following equation for h ' ( x ) first minutes., differentiating the function y = √ ( x4 – 37 ) 12 the! Access now — it ’ s needed is a rule for partial derivatives 1 affiliated!