InterviewSolution
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InterviewSolution
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Differentiate w.r.t. x : `(i)cot^(-1)((1)/(x))" "(ii)tan^(-1)((2x)/(1-x^(2)))" "(iii)cot^(-1)((1-x)/(1+x))` |
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Answer» (i) Let `y=cot^(-1)((1)/(x)).` Putting `x= tan theta`, we get `y=cot^(-1)((1)/(tan theta))=cot^(-1)(cot theta)=theta=tan^(-1)x.` `therefore(dy)/(dx)=(1)/((1+x^(2))).` Hence, `(d)/(dx){cot^(-1)((1)/(2))}=(1)/((1+x^(2))).` (ii) Let `y=tan^(-1)((2x)/(1-x^(2))).` Putting `x=tan theta`, we get `y=tan^(-1)((2tan theta)/(1-tan^(2)theta))=tan^(-1)(tan2 theta)=2theta=2 tan^(-1)x.` `therefore(dy)/(dx)=(2)/((1+x^(2))).` Hence, `(d)/(dx){tan^(-1)((2x)/(1-x^(2)))}=(2)/((1+x^(2))).` (iii) Let `y=cot^(-1)((1-x)/(1+x)).` Putting `x=tan theta,` we get `y=cot^(-1)((1-tan theta)/(1+tan theta))=cot^(-1){tan((pi)/(4)-theta)}` `=cot^(-1)[cot{(pi)/(2)-((pi)/(4)-theta)}]=((pi)/(4)+theta)=(pi)/(4)+tan^(-1)x.` `therefore (dy)/(dx)=(1)/((1+x^(2)))` Hence, `(d)/(dx){cot^(-1)((1-x)/(1+x))}=(1)/((1+x^(2))).` |
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