## I Do not remember how solve this........

Apr 6, 2017

I tried changing the $\csc$:

#### Explanation:

I would use the Quotient and Chain Rule and remember that $\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$ so:
$f \left(x\right) = {\csc}^{2} \left(7 {x}^{2}\right) = \frac{1}{\sin} ^ 2 \left(7 {x}^{2}\right)$
So:
f'(x)=-(2sin(7x^2) cos(7x^2)*14x)/(sin^4(7x^2))=-28x(cos(7x^2)/sin^3(7x^2))=-28xcsc^2(7x^2)cot(7x^2)

Apr 8, 2017

Here I used Chain Rule successively:-

#### Explanation:

$f \left(x\right)$ = ${\csc}^{2} \left(7 {x}^{2}\right)$
Now Differentiating both sides with respect to $x$ successively we get:-
$f ' \left(x\right)$ = 2 $\csc \left(7 {x}^{2}\right) . \left[\frac{d}{\mathrm{dx}} \left(\csc 7 {x}^{2}\right)\right]$
$\Rightarrow f ' \left(x\right)$ = $2 \csc \left(7 {x}^{2}\right) . \left[- \csc \left(7 {x}^{2}\right) . \cot \left(7 {x}^{2}\right) . \frac{d}{\mathrm{dx}} \left(7 {x}^{2}\right)\right]$

$\Rightarrow f ' \left(x\right)$ = $- 2 {\csc}^{2} \left(7 {x}^{2}\right) . \cot \left(7 {x}^{2}\right) .14 x$

$\Rightarrow f ' \left(x\right)$ = $- 28 x . {\csc}^{2} \left(7 {x}^{2}\right) . \cot \left(7 {x}^{2}\right)$