Use the Chain Rule of Differentiation in Calculus

The chain rule of differentiation of functions in calculus is presented along with several examples and detailed solutions and comments.

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 Let $f(x) = (g \circ h)(x) = g(h(x))$ and $u = g(x)$ Let $y = f(u)$; the derivative of $f$ with respect to $x$, $f \, '$ is given by: $f \, '(x) = \dfrac{dy}{du} \dfrac{du}{dx}$ We now present several examples of applications of the chain rule. Example 1: Find the derivative $f \, '(x)$, if $f$ is given by $f(x) = 4 cos (5x - 2)$ Solution to Example 1 Let $u = 5x - 2$ and $y = 4 cos u$, hence   $\dfrac{du}{dx} = 5$ and $\dfrac{dy}{du}=-4 \sin u$ We now use the chain rule   $f \, '(x) = \dfrac{dy}{du} \dfrac{du}{dx} = - 4 \sin (u) (5)$ We now substitute $u = 5 x - 2$ in $sin (u)$ above to obtain   $f \, '(x) = - 20 \sin (5 x - 2)$ Example 2: Find the derivative $f \, '(x)$, if $f$ is given by $f(x) = (x^3 - 4 x + 5)^4$ Solution to Example 2 Let $u = x^3 - 4 x + 5$ and $y = u^4$ , hence   $\dfrac{du}{dx} = 3 x^2 - 4$ and $\dfrac{dy}{du} = 4 u^3$ Use the chain rule   $f \, '(x) = \dfrac{dy}{du}\dfrac{du}{dx} = (4 u^3) (3 x^2 - 4)$ We now substitute $u = x^3 - 4 x + 5$ above to obtain   $f \, '(x) = 4 (x^3 - 4 x + 5)^3 (3 x^2 - 4)$ Example 3: Find $f \, '(x)$, if $f$ is given by $f(x) = \sqrt {x^2 + 2 x -1}$ Solution to Example 3 Let $u = x^2 + 2 x -1$ and $y = \sqrt {u}$, hence   $\dfrac{du}{dx}=2 x + 2$ and $\dfrac{dy}{du}=\dfrac{1}{2\sqrt{u}}$ Use the chain rule   $f \, '(x) = \dfrac{dy}{du} \dfrac{du}{dx} = \dfrac{1}{2 \sqrt{u}} \, (2 x + 2)$ Substitute $u = x^2 + 2 x -1$ above to obtain   $f \, '(x) = (2x + 2) \dfrac{1}{2 \sqrt {x^2 + 2x -1}}$ Factor 2 in numerator and denominator and simplify   $f \, '(x) = \dfrac{x + 1}{\sqrt{x^2 + 2x -1}}$ Example 4: Find the first derivative of function $f$ given by $f(x) = \sin ^2 (2 x + 3)$ Solution to Example 4 Let $u = \sin (2 x + 3)$ and $y = u^2$ , hence   $\dfrac{du}{dx}=2 \cos(2x+3)$ and $\dfrac{dy}{du}=2 u$ Use the chain rule   $f \, '(x) = \dfrac{dy}{du} \dfrac{du}{dx} = 2 u \, 2 \cos(2 x + 3)$ Substitute $u = \sin (2 x + 3)$ above to obtain   $f \, '(x) = 4 \sin (2 x + 3) cos (2 x + 3)$ Use the trigonometric fromula $\sin (2 x) = 2 \sin x \cos x$ to simplify $f \, '(x)$   $f \, '(x) = 2 \sin (4 x + 6)$ Exercises: Find the first derivative to each of the functions. 1 - $f(x) = \cos (3 x -3)$ 2 - $l(x) = (3x^2 -3 x + 8)^4$ 3 - $m(x) = \sin \dfrac{1}{x-2}$ 4 - $t(x) = \sqrt{ 3 x^2 - 3 x + 6}$ 5 - $r(x) = \sin^2 (4 x + 20)$ Solutions to the above exercises 1 - $f \, '(x) = -3 \sin (3x -3)$ 2 - $l \, '(x) = 12 (2x - 1) (3x^2 -3 x + 8)^3$ 3 - $m \, '(x) = -\dfrac{1}{(x - 2)^2} cos \dfrac{1}{x-2}$ 4 - $t\, '(x) = \dfrac{(3/2)(2x - 1)}{\sqrt (3x^2 - 3x + 6)}$ 5 -$r\, '(x) = 8\cos(4x+20)\sin(4x+20) = 4 \sin (8 x + 40)$ More on differentiation and derivatives Solve Rate of Change Problems in Calculus Solve Tangent Lines Problems in Calculus

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Updated: 2 April 2013