Question 587797
y=(sinx)^2(tanx)^4/(x^2+3)^2

again i can't figure it out. any help would be great 
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What do you want to do?
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y=(sinx)^2(tanx)^4*(x^2+3)^-2
Use the product rule:
y' = {{{2sin(x)*cos(x)*tan^4(x)*(x^2+3)^-2 + sin^2(x)*4tan^3(x)*sec^2(x)*(x^2+3)^-2 + sin^2(x)*tan^4(x)*(-2)*(x^2+3)^-3*2x}}}
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Sometimes you can combine and simplify
= {{{2sin(x)*cos(x)*tan^4(x)*(x^2+3)^-2 + 4tan^5(x)*(x^2+3)^-2 -4x*sin^2(x)*tan^4(x)*(x^2+3)^-3}}}
= {{{tan^4(x)*(2sin(x)*cos(x)*(x^2+3)^-2 + 4tan(x)*(x^2+3)^-2 -4x*sin^2(x)*(x^2+3)^-3)}}}
= {{{tan^4(x)*(2sin(x)*cos(x)*(x^2+3) + 4tan(x)*(x^2+3) -4x*sin^2(x))/(x^2+3)^3}}}