AP Calculus AB / Differentiation Composite, Implicit, and Inverse Functions / Question No. 46

46. If $f (x) = {(1 + \frac{x}{20})}^{5},$ find $f^{″} (40)$ .

0.068

1.350

5.400

6.750

Answer is: option2

1.350

Solution:

Using the chain rule, differentiate $f (x)$ : $f^{'} (x) = 5 {(1 + \frac{x}{20})}^{4} \cdot \frac{1}{20}$ $f^{'} (x) = \frac{5}{20} {(1 + \frac{x}{20})}^{4} = \frac{1}{4} {(1 + \frac{x}{20})}^{4}$ Differentiate again: $f^{″} (x) = \frac{1}{4} \cdot 4 {(1 + \frac{x}{20})}^{3} \cdot \frac{1}{20}$ $f^{″} (x) = \frac{1}{20} {(1 + \frac{x}{20})}^{3}$ Substituting $x = 40$ : $f^{″} (40) = \frac{1}{20} {(1 + \frac{40}{20})}^{3}$ $= \frac{1}{20} (1 + 2)^{3}$ $= \frac{1}{20} (3)^{3}$ $= \frac{27}{20}$ $= 1.35$ Final Answer: 1.350

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