Question

# Find derivative of tanâˆ’1cosxâˆ’sinxcosx+sinx w.r.t. x.

A
1
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B
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C
1
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D
2
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Solution

## The correct option is A −1We have to find the derivative of tan−1cosx−sinxcosx+sinx with respect to x.ddx(tan−1cosx−sinxcosx+sinx)Let u=cosx−sinxcosx+sinx,f=tan−1(u)Apply chain rule df(u)dx=dfdu⋅dudx, we getdf(u)dx=ddu(tan−1(u))⋅ddx(cosx−sinxcosx+sinx) =1u2+1⋅(cosx+sinx)(−sinx−cosx)−(cosx−sinx)(−sinx+cosx)(cosx+sinx)2 =1u2+1⋅−(cosx+sinx)2−(cosx−sinx)2(cosx+sinx)2 =1u2+1⋅−cos2x−sin2x−2cosxsinx−cos2x−sin2x+2cosxsinxcos2x+sin2x+2sinxcosx =1u2+1⋅−2cos2x−2sin2x1+2sinxcosx =1u2+1⋅−2(cos2x+sin2x)1+2sinxcosx =1u2+1⋅−21+2sinxcosxSubstitute u=cosx−sinxcosx+sinx we get =1(cosx−sinxcosx+sinx)2+1⋅−21+2sinxcosx =(cosx+sinx)2(cosx−sinx)2+(cosx+sinx)2⋅−21+2sinxcosx =cos2x+sin2x+2cosxsinxcos2x+sin2x−2cosxsinx+cos2x+sin2x+2sinxcosx⋅−21+2sinxcosx =1+2cosxsinx1−2cosxsinx+1+2sinxcosx⋅−21+2sinxcosx =1+2sinxcosx2⋅−21+2sinxcosx =−1ddx(tan−1cosx−sinxcosx+sinx)=−1

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