How do you find the derivative of ${cos}^{2} x sin x$ $cos^2x sinx$ ?

Question

9663 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-27T18:09:40

$"the derivative of y="cos^2 x sin x" is "y^'=-2cos x+3cos^3 x$

$a=cos ^2x$

$b=sin x$

$y=cos^2 x sin x$

$y=a*b$

$y^'=a^'*b+b^'*a$

$a^'=-2cos x*sin x$

$b^'=cos x$

$y^'=-2cos x*sin x* sin x+cos x*cos^2 x$

$y^'=-2cos x*sin^2 x+cos^3 x$

$sin^2 x=1-cos^2 x$

$y^'=-2cos x(1-cos^2 x)+cos^3 x$

$y^'=-2cos x+2cos^3 x+cos^3 x$

$y^'=-2cos x+3cos^3 x$

Answer 2 · 2016-11-27T18:45:14

I would replace $cos^2x$ by $1-sin^2x$

$cos^2xsinx = (1-sin^2x)sinx = sinx-sin^3x$

The derivative is $cosx-3sin^2xcosx$ .

There are other ways to write the derivative.

How do you find the derivative of cos^2x sinxcos2xsinxcos^2x sinx?