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I have simple problem in derivative of trigonometric functions, can anyone show me the way to calculate the derivative of following function?

$$y = \sin^2 x (\cos x + \sin x)$$

I think the answer should be:

$$y' = (2 \sin x \cos x)(\cos x + \sin x) + (-\sin x + \cos x)(\sin^2 x)$$

Is the above answer correct?

Thanks in Advance...

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  • $\begingroup$ You should apply the chain rule to differentiate $\sin^2 x$. $\endgroup$ – David Mitra Jan 4 '14 at 16:18
  • $\begingroup$ It's incorrect. What's the derivative of $x\mapsto (\sin(x))^2$? $\endgroup$ – Git Gud Jan 4 '14 at 16:18
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No. $$(\sin^2 x)^\prime\not=2\sin x.$$ $$(\sin^2 x)^\prime=2\sin x\cdot (\sin x)^\prime=2\sin x\cos x.$$

The other parts are fine.

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$$\frac{d\{\sin^2x(\cos x+\sin x)\}}{dx}=(\cos x+\sin x)\frac{d(\sin^2x)}{dx}+\sin^2x\frac{d(\cos x+\sin x)}{dx}$$

$$=(\cos x+\sin x)\cdot\frac{d(\sin^2x)}{d(\sin x)}\cdot\frac{d(\sin x)}{dx}+\sin^2x(-\cos x+\cos x)$$

$$=\cdots$$

Reference : Product rule and Chain Rule

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