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Question

Find derivative of tan1cosxsinxcosx+sinx w.r.t. x.

A
1
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B
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C
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2
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Solution

The correct option is A 1
We have to find the derivative of tan1cosxsinxcosx+sinx with respect to x.

ddx(tan1cosxsinxcosx+sinx)

Let u=cosxsinxcosx+sinx,f=tan1(u)

Apply chain rule df(u)dx=dfdududx, we get

df(u)dx=ddu(tan1(u))ddx(cosxsinxcosx+sinx)

=1u2+1(cosx+sinx)(sinxcosx)(cosxsinx)(sinx+cosx)(cosx+sinx)2

=1u2+1(cosx+sinx)2(cosxsinx)2(cosx+sinx)2

=1u2+1cos2xsin2x2cosxsinxcos2xsin2x+2cosxsinxcos2x+sin2x+2sinxcosx

=1u2+12cos2x2sin2x1+2sinxcosx

=1u2+12(cos2x+sin2x)1+2sinxcosx

=1u2+121+2sinxcosx

Substitute u=cosxsinxcosx+sinx we get

=1(cosxsinxcosx+sinx)2+121+2sinxcosx

=(cosx+sinx)2(cosxsinx)2+(cosx+sinx)221+2sinxcosx

=cos2x+sin2x+2cosxsinxcos2x+sin2x2cosxsinx+cos2x+sin2x+2sinxcosx21+2sinxcosx

=1+2cosxsinx12cosxsinx+1+2sinxcosx21+2sinxcosx

=1+2sinxcosx221+2sinxcosx

=1

ddx(tan1cosxsinxcosx+sinx)=1


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