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Question: What is the derivative of \(\sin x(\sin x + \cos x)\)?...

What is the derivative of sinx(sinx+cosx)\sin x(\sin x + \cos x)?

Explanation

Solution

We use Product – Rule to find the derivative of function sinx(sinx+cosx)\sin x(\sin x + \cos x).
The product rule helps us to differentiate between two or more of the functions in a given function.
If uu and vv are the two given function of xx then the product rule is given by the following formula:
d(uv)dx=udvdx+vdudx\dfrac{{d(uv)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}. After product rule we use different trigonometric identities such as cos2x=cos2xsin2x\cos 2x = {\cos ^2}x - {\sin ^2}x and sin2x=2sinxcosx\sin 2x = 2\sin x\cos x to solve the given problem.

Complete step by step answer:
The given function is the product of two functions sinx\sin x and (sinx+cosx)(\sin x + \cos x).
We use product rule to find the derivative, that means first we multiply the first function by the derivative of the second function and the second function is multiplied by the derivative of the first function and add them.
So, we can write it as
sinx(d(sinx+cosx)dx)+(sinx+cosx)(d(sinx)dx)(1)\Rightarrow \sin x\left( {\dfrac{{d(\sin x + \cos x)}}{{dx}}} \right) + (\sin x + \cos x)\left( {\dfrac{{d(\sin x)}}{{dx}}} \right)\,\, \ldots \ldots \ldots (1)
We know that the derivative of sinx=cosx\sin x = \cos x and the derivative of cosx=sinx\cos x = - \sin x.
So, d(sinx+cosx)dx=(cosxsinx)\dfrac{{d(\sin x + \cos x)}}{{dx}} = (\cos x - \sin x) and dsinxdx=cosx\dfrac{{d\sin x}}{{dx}} = \cos x
Put these values in equation 11. We get,
sinx(cosxsinx)+(sinx+cosx)(cosx)\Rightarrow \sin x(\cos x - \sin x) + (\sin x + \cos x)(\cos x)
Simplifying the above equation. We get,
sinxcosxsin2x+sinxcosx+cos2x\Rightarrow \sin x\cos x - {\sin ^2}x + \sin x\cos x + {\cos ^2}x
2sinxcosx+cos2xsin2x\Rightarrow 2\sin x\cos x + {\cos ^2}x - {\sin ^2}x
We know that cos2x=cos2xsin2x\cos 2x = {\cos ^2}x - {\sin ^2}x and sin2x=2sinxcosx\sin 2x = 2\sin x\cos x.
Write these values in the above equation. We get,
sin2x+cos2x\Rightarrow \sin 2x + \cos 2x
Hence, the derivative of the function sinx(sinx+cosx)\sin x(\sin x + \cos x) is sin2x+cos2x\sin 2x + \cos 2x.

Note:
To find the derivative of the given function we can also use chain rule. First we simplify the expression by opening brackets and we will get a composite function. Chain rule is used where the function is composite, we can denote chain rule by f.gf.g, where ff and gg are two functions. Chain rule states that the derivative of a composite function can be taken as the derivative of the outer function which we multiply by the derivative of the inner function.