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

Question

11217 views around the world

You can reuse this answer
Creative Commons License

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)#.