Question

If f(y) = e^(9y) + e^(-9y), find f'(y). Use exact values.

Answer 1

Given:

$f(y)=e^{9y}+e^{-9y}$

To find:

The f'(y).

Solution:

Chain rule of differentiation:

$[f(g(x))]'=f'(g(x))g'(x)$

Differentiation of exponential:

$\dfrac{d}{dx}e^x=e^x$

We have,

$f(y)=e^{9y}+e^{-9y}$

Differentiate with respect to y.

$f'(y)=e^{9y}\dfrac{d}{dx}(9y)+e^{-9y}\dfrac{d}{dx}(-9y)$

$f'(y)=e^{9y}\cdot 9(1)+e^{-9y}\cdot (-9)(1)$

$f'(y)=9e^{9y}-9e^{-9y}$

Therefore, the differentiation of the given function is $f'(y)=9e^{9y}-9e^{-9y}$.