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3.3 Derivatives of Composite Functions: The Chain …

Derivatives OF Composite functions : THE Chain RULE1. Derivatives of Composite functions : The Chain Rule In this section we want to find the derivative of a Composite function f (g(x)). where f (x) and g(x) are two differentiable functions . Theorem If f and g are differentiable then f (g(x)) is differentiable with derivative given by the formula d f (g(x)) = f 0 (g(x)) g 0 (x). dx This result is known as the Chain rule. Thus, the derivative of f (g(x)) is the derivative of f (x) evaluated at g(x) times the derivative of g(x). Proof. By the definition of the derivative we have d f (g(x + h)) f (g(x)). f (g(x)) = lim . dx h 0 h Since g is differentiable at x, letting g(x + h) g(x). v= g 0 (x). h we find g(x + h) = g(x) + (v + g 0 (x))h with limh 0 v = 0. Similarly, we can write f (y + k) = f (y) + (w + f 0 (y))k with limk 0 w = 0.

3.3 DERIVATIVES OF COMPOSITE FUNCTIONS: THE CHAIN RULE1 3.3 Derivatives of Composite Functions: The Chain Rule In this section we want to nd the derivative of a composite function f(g(x))

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Transcription of 3.3 Derivatives of Composite Functions: The Chain …

1 Derivatives OF Composite functions : THE Chain RULE1. Derivatives of Composite functions : The Chain Rule In this section we want to find the derivative of a Composite function f (g(x)). where f (x) and g(x) are two differentiable functions . Theorem If f and g are differentiable then f (g(x)) is differentiable with derivative given by the formula d f (g(x)) = f 0 (g(x)) g 0 (x). dx This result is known as the Chain rule. Thus, the derivative of f (g(x)) is the derivative of f (x) evaluated at g(x) times the derivative of g(x). Proof. By the definition of the derivative we have d f (g(x + h)) f (g(x)). f (g(x)) = lim . dx h 0 h Since g is differentiable at x, letting g(x + h) g(x). v= g 0 (x). h we find g(x + h) = g(x) + (v + g 0 (x))h with limh 0 v = 0. Similarly, we can write f (y + k) = f (y) + (w + f 0 (y))k with limk 0 w = 0.

2 In particular, letting y = g(x) and k = (v + g 0 (x))h we find f (g(x) + (v + g 0 (x))h) = f (g(x)) + (w + f 0 (g(x)))(v + g 0 (x))h. Hence, f (g(x + h)) f (g(x)) = f (g(x) + (v + g 0 (x))h) f (g(x)). = f (g(x)) + (w + f 0 (g(x)))(v + g 0 (x))h f (g(x)). = (w + f 0 (g(x)))(v + g 0 (x))h 2. Thus, d f (g(x + h)) f (g(x)). f (g(x)) = lim dx h 0 h = lim (w + f 0 (g(x)))(v + g 0 (x)). h 0. = f (g(x))g 0 (x). 0. This completes a proof of the theorem Example Find the derivative of y = (4x2 + 1)7 . Solution. First note that y = f (g(x)) where f (x) = x7 and g(x) = 4x2 + 1. Thus, f 0 (x) = 7x6 , f 0 (g(x)) = 7(4x2 + 1)6 and g 0 (x) = 8x. So according to the Chain rule, y 0 = 7(4x2 + 1)6 (8x) = 56x(4x2 + 1)6. Example Prove the power rule for rational exponents. Solution. p Suppose that y = x q , where p and q are integers with q > 0.

3 Take the q th power of both sides to obtain y q = xp . Differentiate both sides with respect to x to obtain qy q 1 y 0 = pxp 1 . Thus, p xp 1 p p y0 = p(q 1). = x q 1 . qx q q p Note that we are assuming that x is chosen in such a way that x q is defined Example d n Show that dx x = nxn 1 for x > 0 and n is any real number. Solution. Since xn = en ln x then d n d n ln x n x = e = en ln x = nxn 1 . dx dx x Derivatives OF Composite functions : THE Chain RULE3. We end this section by finding the derivative of f (x) = ln x using the Chain rule. Write y = ln x. Then ey = x. Differentiate both sides with respect to x to obtain ey y 0 = 1. Solving for y 0 we find 1 1. y0 = y = . e x


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