What is the derivative of $\frac{{sin}^{2} x}{cos x}$ $sin^2x/cosx$ ?

Question

4391 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-28T01:03:03

$f'(x)=((sinx)(2cos^2x+sin^2x))/cos^2x$

$f(x)=(sin^2x)/cosx$

$f'(x)=(cosxtimes2sinxcosx-sin^2xtimes-sinx)/cos^2x$

$f'(x)= (2sinxcos^2x+sin^3x)/cos^2x$

$f'(x)=((sinx)(2cos^2x+sin^2x))/cos^2x$

The quotient rule is given by:

$f(x)=u/v$

$f'(x)=(vu'-uv')/v^2$

What is the derivative of sin^2x/cosxsin2xcosxsin^2x/cosx?