Question

Consider the function $f^{\prime \prime }\left(x\right)={\sec }^{4}x+4$ with $f\left(0\right)=0$ and $f^{\prime }\left(0\right)=0$. What is $f^{\prime }\left(x\right)$ equal to?

• A. $\tan x-\frac{{\tan }^{3}x}{3}+4x$
• B. $\tan x+\frac{{\tan }^{3}x}{3}+4x$
• C. $\tan x+\frac{{\sec }^{3}x}{3}+4x$
• D. $-\tan x-\frac{{\tan }^{3}x}{3}+4x$

Answer 1

The correct option is B $\tan x+\frac{{\tan }^{3}x}{3}+4x$

Given that $f^{\prime \prime }\left(x\right)={\sec }^{2}x+4$

Integrating on both sides, we get $f^{\prime }\left(x\right)=\int ({\sec }^{4}x+4)dx$

$=\int {\sec }^{2}x(1+{\tan }^{2}x)dx+4x+c$

$=\int ({\sec }^{2}x+{\sec }^{2}x{\tan }^{2}x)dx+4x+c$

$=\int {\sec }^{2}xdx+\int {\sec }^{2}x{\tan }^{2}xdx+4x+c$

$=\tan x+\frac{{\tan }^{3}x}{3}+4x+c$

substitute $f^{\prime }\left(0\right)=0$ we get $c=0$

$f^{\prime }\left(x\right)=\tan x+\frac{{\tan }^{3}x}{3}+4x$