Calculus (3rd Edition)

Published by W. H. Freeman

Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 37

Answer

$$f'(x) =\frac{(3x-2)e^x}{(3x+1)^2}$$

Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(e^x)'=e^x$ Since we have $$f(x)= \frac{e^x}{3x+1}$$ then the derivative $f'(x)$, using the chain and quotient rules, is given by $$f'(x)=\frac{(3x+1)(e^x)'-e^x(3x+1)'}{(3x+1)^2} =\frac{(3x+1)e^x-3e^x}{(3x+1)^2}\\ =\frac{(3x-2)e^x}{(3x+1)^2}$$

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