Find the Derivative (dy/dx) of the following problem:

Question

livya15 · Answer

$$y=\cot ^{2} 5x \sin5x $$

anonymous · Answer

$$\frac{ dy }{ dx }=2\cot 5x \left( -\csc ^25x \right)*5\sin 5x+2\cot ^25x*\cos 5x*5$$
$$=10\cot 5x \sin 5x \left( -\csc ^25x+\cot 5x \right)$$
?

livya15 · Answer

What happened to the cot5x and the cot^2 5x?
@surjithayer

anonymous · Answer

Did you get first step?

anonymous · Answer

$$\frac{ dy }{ dx }=2\cot 5x \left( -cosec ^25x \right) \cdot5\sin 5x+2\cot ^25x \cdot \cos 5x \cdot 5$$

livya15 · Answer

I understand that step, i am just confused of how you simplify it...

phi · Answer

I would simplify first, then find the derivative
cot  = cos/sin

and cot^2 5x  * sin 5x =  cos^2 5x / sin^2 5x  * sin 5x =cos^2 5x / sin 5x
= (1- sin^2 5x) / sin 5x =  1/ sin 5x  - sin 5x
now take the derivative.

livya15 · Answer

thanks