Question

Find the derivative y=x^2csc5x.

Answer 1

\(y = \csc x\) then \(y' = -\csc x \cot x\)
Use that and product rule and chain rule.

Answer 2

and power rule.

Answer 3

Product rule:
\( y = uv\) then \(y' = uv' + vu'\)

Answer 4

\(y=x^2\csc5x\)

\(y' = x^2(\csc 5x)' + \csc 5x(x^2)'\)

Answer 5

Can you finish it now?

Answer 6

Sure. Give me a second.

Answer 7

Is it x(2-5xcot(5 x))csc(5 x)?

Answer 8

\(y' = x^2 (-5\csc 5x \cot 5x) + \csc 5x (2x)\)

\(y' =-5x^2 \csc 5x \cot 5x + 2x \csc 5x\)

\(y' = x \csc 5x (2-5x \cot 5x)\)

Answer 9

Yes, that's the same I got.

Answer 10

Oh ok. Thanks!

Answer 11

You're welcome.