Question

if $f(x)=\frac{4x^{2}\tan x}{sec x}$, find $f(x)=$
find $f(1)=$

Explanation:

Step1: Simplify the function

First, recall that $\frac{\tan x}{\sec x}=\sin x$. So $f(x) = 4x^{2}\sin x$.

Step2: Find the first - derivative using product rule

The product rule is $(uv)'=u'v + uv'$, where $u = 4x^{2}$ and $v=\sin x$. $u'=8x$ and $v'=\cos x$. Then $f'(x)=8x\sin x+4x^{2}\cos x$.

Step3: Find the second - derivative using product rule again

For the first term $8x\sin x$: Let $u = 8x$ and $v=\sin x$, then $(8x\sin x)'=8\sin x + 8x\cos x$. For the second term $4x^{2}\cos x$: Let $u = 4x^{2}$ and $v=\cos x$, then $(4x^{2}\cos x)'=8x\cos x-4x^{2}\sin x$. So $f''(x)=8\sin x + 8x\cos x+8x\cos x-4x^{2}\sin x=8\sin x + 16x\cos x-4x^{2}\sin x$.

Step4: Evaluate $f''(1)$

Substitute $x = 1$ into $f''(x)$: $f''(1)=8\sin(1)+16\cos(1)-4\sin(1)=4\sin(1)+16\cos(1)$.

Answer:

$f''(x)=8\sin x + 16x\cos x-4x^{2}\sin x$; $f''(1)=4\sin(1)+16\cos(1)$