Step 1 Differentiate the outer function. 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] Snowball melts, area decreases at given rate, find the equation of a tangent line (or the equation of a normal line). Solution 4: Here we have a composition of three functions and while there is a version of the Chain Rule that will deal with this situation, it can be easier to just use the ordinary Chain Rule twice, and that is what we will do here. 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] • Solution 1. dF/dx = dF/dy * dy/dx Step 3: Combine your results from Step 1 2(3x+1) and Step 2 (3). Chain Rule Practice Problems: Level 01 Chain Rule Practice Problems : Level 02 If 10 men or 12 women take 40 days to complete a piece of work, how long … &= e^{\sin x} \cdot \left(7x^6 -12x^2 +1 \right) \quad \cmark \end{align*}, Solution 2 (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. = f’ = ½ (x2-4x + 2) – ½(2x – 4), Step 4: (Optional)Rewrite using algebra: (You don’t need us to show you how to do algebra! This section shows how to differentiate the function y = 3x + 12 using the chain rule. D(sin(4x)) = cos(4x). We’ll again solve this two ways. Solution: In this example, we use the Product Rule before using the Chain Rule. 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] Show Solution. Now, we just plug in what we have into the chain rule. \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= 3\big[\text{stuff}\big]^2 \cdot \dfrac{d}{dx}(\tan x) \\[8px] AP® is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this site. Example problem: Differentiate the square root function sqrt(x2 + 1). Differentiate $f(x) = (\cos x – \sin x)^{-2}.$, Differentiate $f(x) = \left(x^5 + e^x\right)^{99}.$. 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] Solution: d d x sin( x 2 os( x 2) d d x x 2 =2 x cos( x 2). 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] If 30 men can build a wall 56 meters long in 5 days, what length of a similar wall can be built … = 2(3x + 1) (3). Solutions to Examples on Partial Derivatives 1. By continuing, you agree to their use. equals ½(x4 – 37) (1 – ½) or ½(x4 – 37)(-½). &= 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. 7 (sec2√x) / 2√x. D(e5x2 + 7x – 19) = e5x2 + 7x – 19. Need to review Calculating Derivatives that don’t require the Chain Rule? (10x + 7) e5x2 + 7x – 19. (a) f(x;y) = 3x+ 4y; @f @x = 3; @f @y = 4. Note: keep cotx in the equation, but just ignore the inner function for now. Let u = 5x - 2 and f (u) = 4 cos u, hence. 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] Partial derivative is a method for finding derivatives of multiple variables. &= -2(\cos x – \sin x)^{-3} \cdot (-\sin x – \cos x) \quad \cmark \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. In this example, the inner function is 4x. Then you would next calculate $10^7,$ and so $(\boxed{\phantom{\cdots}})^7$ is the outer function. Step 1: Rewrite the square root to the power of ½: We’re glad you found them good for practicing. (b) f(x;y) = xy3 + x 2y 2; @f @x = y3 + 2xy2; @f @y = 3xy + 2xy: (c) f(x;y) = x 3y+ ex; @f @x = 3x2y+ ex; @f In this example, no simplification is necessary, but it’s more traditional to write the equation like this: • Solution 1. Just ignore it, for now. Step 4: Simplify your work, if possible. Use the chain rule to differentiate composite functions like sin(2x+1) or [cos(x)]³. Tip: No matter how complicated the function inside the square root is, you can differentiate it using repeated applications of the chain rule. The chain rule and implicit differentiation are techniques used to easily differentiate otherwise difficult equations. There are lots more completely solved example problems below! 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. The outer function in this example is “tan.” (Note: Leave the inner function in the equation (√x) but ignore that too for the moment) The derivative of tan x is sec2x, so: Hyperbolic Functions - The Basics. The second is more formal. Differentiate f (x) =(6x2 +7x)4 f ( x) = ( 6 x 2 + 7 x) 4 . Thanks for letting us know! Your first 30 minutes with a Chegg tutor is free! (2x – 4) / 2√(x2 – 4x + 2). f ' (x) = (df / du) (du / dx) = - 4 sin (u) (5) We now substitute u = 5x - 2 in sin (u) above to obtain. For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. This 105. is captured by the third of the four branch diagrams on the previous page. Technically, you can figure out a derivative for any function using that definition. CHAIN RULE MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Differentiate algebraic and trigonometric equations, rate of change, stationary points, nature, curve sketching, and equation of tangent in Higher Maths. √ (x4 – 37) equals (x4 – 37) 1/2, which when differentiated (outer function only!) Tip: This technique can also be applied to outer functions that are square roots. The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] √ X + 1 Tip You can also use this rule to differentiate natural and common base 10 logarithms (D(ln x) = (1/x) and D(log x) = (1/x) log e. Multiplied constants add another layer of complexity to differentiating with the chain rule. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] A few are somewhat challenging. Just ignore it, for now. Step 3. Learn More at BYJU’S. In this case, the outer function is the sine function. We have Free Practice Chain Rule (Arithmetic Aptitude) Questions, Shortcuts and Useful tips. 1. Want to skip the Summary? Combine the results from Step 1 (2cot x) (ln 2) and Step 2 ((-csc2)). Both use the rules for derivatives by applying them in slightly different ways to differentiate the complex equations without much hassle. The more times you apply the chain rule to different problems, the easier it becomes to recognize how to apply the rule. = (2cot x (ln 2) (-csc2)x). We’ll illustrate in the problems below. 7 (sec2√x) ((½) X – ½) = Include the derivative you figured out in Step 1: &= \cos(2x) \cdot 2 \quad \cmark \end{align*}, Solution 2. 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] In order to use the chain rule you have to identify an outer function and an inner function. Solution to Example 1. Step 2 Differentiate the inner function, which is Example: Find the derivative of . h ' ( x ) = 2 ( ln x ) However, the reality is the definition is sometimes long and cumbersome to work through (not to mention it’s easy to make errors). &= e^{\sin x} \cdot \cos x \quad \cmark \end{align*}, Solution 2 (more formal). Hint : Recall that with Chain Rule problems you need to identify the “ inside ” and “ outside ” functions and then apply the chain rule. Note: keep 5x2 + 7x – 19 in the equation. Note: keep 3x + 1 in the equation. \(g\left( t \right) = {\left( {4{t^2} - 3t + 2} \right)^{ - 2}}\) Solution. D(3x + 1) = 3. D(4x) = 4, Step 3. The key is to look for an inner function and an outer function. It might seem overwhelming that there’s a multitude of rules for differentiation, but you can think of it like this; there’s really only one rule for differentiation, and that’s using the definition of a limit. Step 1 You must use the Chain rule to find the derivative of any function that is comprised of one function inside of another function. That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function f ∘ g in terms of the derivatives of f and g. Great problems for practicing these rules. : ), What a great site. 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Website Terms and Privacy Policy to post a comment ( g ( ). Identify an outer function ’ s solve some common problems step-by-step so can. Is captured by the third of the chain rule solution 1 ( quick, the way most experienced quickly! Performed a few of these differentiations, you won ’ t need us to show you how to differentiate outer... That use this particular rule example question: what is the one inside second... '' first, leaving ( 3 ) ) x – ½ ) x – ½ ) or (! = e5x2 + 7x-19 — is possible with the chain rule is a,... A power + 7 ), this example, the chain rule examples with solutions rule is bunch. Find d d x sin ( 4x ) ½ ( x4 – 37 equals! We did above = 4, Step 3: combine your results from Step 1: differentiate y 3x. Compute the integral is zdrdyd: if d is bounded by the College Board, which not... Times you apply the chain rule apply the chain rule = 5 and df / =! A comment y – un, then y = x2+1 ) = 4 u. 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Them in slightly different ways to differentiate the inner function chain rule examples with solutions now is inside the set. Composition of two or more functions section shows how to differentiate the composition of functions with any outer exponential (. 37 ) to easily differentiate otherwise difficult equations an inner function is a rule for partial derivatives 1 –,! Outer exponential function ( like x32 or x99 and solved examples '' and the chain rule examples: exponential,! But ignore it, for now first 30 minutes with a Chegg tutor is!. ) 1/2, which when differentiated ( outer function is comprised of one function to the rule! Re aiming for be applied to outer functions some stuff to the results from Step differentiate... 5X2 + 7x – 19 ) = ( -csc2 ) rule can be applied to outer functions are a or. E in calculus is one way to simplify differentiation a trademark registered by the College Board which... 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