InterviewSolution
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InterviewSolution
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Differentiate w.r.t. x: `(i)cos^(-1)(4x^(3)-3x)" "(ii)sin^(-1)((1-x^(2))/(1+x^(2)))" "(iii)sec^(-1)((x^(2)+1)/(x^(2)-1))` |
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Answer» (i) Let `y=cos^(-1)(4x^(3)-3x).` Putting `x=cos theta`, we get `y=cos^(-1)(4cos^(3) theta-3 cos theta)=cos^(-1)(cos 3 theta)=3 theta.` `therefore y=3 theta rArr y= 3 cos^(-1)x` `rArr (dy)/(dx)=((-3)/(sqrt(1-x^(2))))`. Let `y=sin^(-1)((1-x^(2))/(1+x^(2)))`. Putting `x=tan theta,` we get `y=sin^(-1)((1-tan^(2)theta)/(1+tan^(2)theta))=sin^(-1)(cos2 theta)=sin^(-1)[sin((pi)/(2)-2 theta)]` `=((pi)/(2)-2 theta)=((pi)/(2)-2 tan^(-1)x).` `therefore y=((pi)/(2)-2 tan^(-1)x).` Hence, `(dy)/(dx)=(d)/(dx)((pi)/(2)-2 tan^(-1)x)=(d)/(dx)((pi)/(2))-2.(d)/(dx)(tan^(-1)x)` `={0-(2)/((1+x^(2)))}=(-2)/((1+x^(2))).` Let `y=sec^(-1)((x^(2)+1)/(x^(2)-1)).` Putting `x=cot theta`, we get Putting `x=cot theta`, we get `y=sec^(-1)((cot^(2)theta+1)/(cot^(2) theta-1))=sec^(-1)((1+tan^(2)theta)/(1-tan^(2)theta))` `=sec^(-1)(sec 2 theta)=2 theta=2 cot^(-1)x.` `therefore (dy)/(dx)=(-2)/((1+x^(2)))` Hence, `(d)/(dx){sec^(-1)((x^(2)+1)/(x^(2)-1))}=(-2)/((1+x^(2))).` |
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