If y = tan (cot⁻¹ x), then \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) at x = 1 is equal to

Question

Question

Answer 1

Concept:

\({\cot ^{ - 1}}{\rm{x}} = {\rm{\;}}{\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right)\)

tan (tan^-1 x) = x

Calculation:

Given:

y = tan (cot⁻¹ x)

\( \Rightarrow {\rm{y}} = \tan \left[ {{{\tan }^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right)} \right]\)

⇒ y = 1/x (∵tan (tan^-1 x) = x)

Differentiating with respect to x, we get

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{\;}} - \frac{1}{{{{\rm{x}}^2}}}\)

At x = 1

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \; - \frac{1}{1} = \; - 1\)