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Algebra 20 Online
OpenStudy (anonymous):

What is the derivative of sinx^2?

OpenStudy (anonymous):

\[\sin ^{2}x\] ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

ok can you simplify sin^2 x ?

OpenStudy (anonymous):

1-cos2x/2

OpenStudy (anonymous):

now you have the derivative of 1/2 minus the derivative of -cos2x/2

OpenStudy (anonymous):

the derivative of 1/2 is zero

OpenStudy (anonymous):

and that of -cos2x/2 is 2*sin2x

OpenStudy (anonymous):

thanks))

OpenStudy (anonymous):

anytime :)

OpenStudy (alekos):

I think it should be just sin2x

OpenStudy (akashdeepdeb):

The derivative of \(sin^2 x~~\neq ~~2sin 2x\)

OpenStudy (akashdeepdeb):

It is just sin 2x

OpenStudy (atlas):

you need to use chain rule to solve this: dy/dx = (dy/dt) * (dt/dx)

OpenStudy (atlas):

d(sin^2x)/dx = {d(sin^2x)/d(sinx)} * {d(sinx)/dx} here i let sinx =y so, d(y^2)/dx = {d(y^2)/dy} * {dy/dx} = 2y dy/dx = 2 sinx d(sinx)/dx = 2 sinx cosx = sin2x

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