Derivative of trigonometric function How i can find the derivative of this trigonometric function
$csc^4(8x^4-5)$ 
i tried to do it my self and i got to this 
$ 4[csc(8x^4-5)]^3 * [-csc(8x^4-5)cotan(8x^4-5)] $
The answer in the book is $ -128 x^3csc^4(8x^4-5)cot an(8x^4-5)$

$$\csc^4(8x^4-5)=\frac{1}{\sin^4(8x^4-5)}=-\frac{(\sin^4(8x^4-5))'}{\sin^8(8x^4-5)}=-\frac{4\sin^3(8x^4-5)\cdot (\sin(8x^4-5))'}{\sin^8(8x^4-5)}$$
$$=-\frac{4\sin^3(8x^4-5)\cdot \cos(8x^4-5)(8x^4-5)'}{\sin^8(8x^4-5)}=-\frac{4\sin^3(8x^4-5)\cdot \cos(8x^4-5)\cdot 32x^3}{\sin^8(8x^4-5)}$$
$$=-\frac{128 x^3\cdot \cos(8x^4-5)}{\sin^5(8x^4-5)}=-128x^3\frac{\cos(8x^4-5)}{\sin(8x^4-5)}\cdot \frac{1}{\sin^4(8x^4-5)}$$
$$=-128x^3\cot(8x^4-5)\csc^4(8x^4-5)$$
