Question

the derivative of tan inverse x with respect to cot inverse x is ​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given:

$tan {}^{ - 1} x \: \: and \: \: {cot}^{ - 1} x$

To find:Find the derivative of tan^-1 x wrt to cot^-1 x

Solution:

Let

$\bold{\red{p = tan {}^{ - 1} x}}$

and

$\bold{\green{q = cot {}^{ - 1} x}}$

Find dp/dx and dq/dx

$\frac{dp}{dx} = \frac{1}{1 + {x}^{2} } \: ...eq1$

and

$\frac{dq}{dx} = - \frac{1}{1 + {x}^{2} } ...eq2$

Divide eq1 by eq2

$\frac{dp}{dq} = \frac{1}{1 + {x}^{2} } \times - \frac{1 + {x}^{2} }{1} \\ \\ \frac{dp}{dq} = - 1$

Final answer:

$\bold{\red{ \frac{dp}{dq} = - 1}}$

Hope it helps you.

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