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What is the derivative of sinx^2?
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\[\sin ^{2}x\] ?
yes
ok can you simplify sin^2 x ?
1-cos2x/2
now you have the derivative of 1/2 minus the derivative of -cos2x/2
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the derivative of 1/2 is zero
and that of -cos2x/2 is 2*sin2x
thanks))
anytime :)
I think it should be just sin2x
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The derivative of \(sin^2 x~~\neq ~~2sin 2x\)
It is just sin 2x
you need to use chain rule to solve this: dy/dx = (dy/dt) * (dt/dx)
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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