How do you find the derivative of $y=sin^2x/cosx$ ?

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Answer 1 · 2016-01-05T02:53:17

$y'=(2cos^2xsinx+sin^3x)/cos^2x$

Use the quotient rule, which states that for a function $y=(f(x))/(g(x))$ ,

$y'=(f'(x)g(x)-g'(x)f(x))/(g(x))^2$

We have:

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

To find $f'(x)$ , use the chain rule: $d/dx(u^2)=2u*u'$ .

$f'(x)=2sinxcosx$
$g'(x)=-sinx$

Plug these into the quotient rule equation.

$y'=(2sinxcosx(cosx)-(-sinx)sin^2x)/(cos^2x)$

$y'=(2cos^2xsinx+sin^3x)/cos^2x$

This alternatively can be simplified as

$y'=sin^3x/cos^2x+2sinx$

Answer 2 · 2016-01-05T03:28:16

$frac{dy}{dx} = sinx(frac{1}{cos^2x}+1)$

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

Therefore,

$y = frac{1 - cos^2x}{cosx}$

$= 1/cosx - cosx$ .

Using the Chain Rule,

$frac{dy}{dx} = frac{d}{dx}(1/cosx) - frac{d}{dx}(cosx)$

$= frac{-1}{cos^2x}(-sinx) + sinx$

$= sinx(frac{1}{cos^2x}+1)$ .

How do you find the derivative of y=sin^2x/cosxy=sin^2x/cosx?