Question

let
$f(x)=-5\cos x + 4\tan x$
$\cdot f(x)=\square$
$\cdot f(-\frac{\pi}{4})=\square$
(1 point)
note: you can earn partial credit on this problem.
submit answers
preview my answers
you have attempted this problem 0 times.
you have unlimited attempts remaining.
email instructor

Explanation:

Step1: Differentiate the function $f(x)$

Use the derivative rules: $\frac{d}{dx}(\cos x)=-\sin x$ and $\frac{d}{dx}(\tan x)=\sec^{2}x$.
For $f(x)= - 5\cos x+4\tan x$, by the sum - rule of derivatives $\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}$ (where $u=-5\cos x$ and $v = 4\tan x$).
The derivative of $u=-5\cos x$ is $u^\prime=-5\times(-\sin x)=5\sin x$, and the derivative of $v = 4\tan x$ is $v^\prime=4\sec^{2}x$. So $f^\prime(x)=5\sin x + 4\sec^{2}x$.

Step2: Evaluate $f^\prime(-\frac{\pi}{4})$

First, recall that $\sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$ and $\sec x=\frac{1}{\cos x}$, so $\sec(-\frac{\pi}{4})=\frac{1}{\cos(-\frac{\pi}{4})}=\sqrt{2}$ (since $\cos(-\alpha)=\cos\alpha$ and $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$).
Substitute $x =-\frac{\pi}{4}$ into $f^\prime(x)$:

$$
LATEXBLOCK0
$$

Answer:

$f^\prime(x)=5\sin x + 4\sec^{2}x$; $f^\prime(-\frac{\pi}{4})=8-\frac{5\sqrt{2}}{2}$