What is the derivative of tan^-1 x with respect to cot^-1 x ?

Concept:

\(\rm \frac{d(\tan^{-1} x)}{dx} = \frac{1}{1+x^2}\)

\(\rm \frac{d(\tan^{-1} x)}{dx} = \frac{1}{1+x^2}\)

Steps for derivatives of functions expressed in the parametric form:

1. First of all, we write the given functions u and v in terms of the parameter x.
2. Using differentiation find out du/dx and dv/dx.
3. Then by using the formula used for solving functions in parametric form i.e.
4.  Lastly substituting the values of du/dx and dv/dx and simplify to obtain the result.

Calculation:

Let u = tan-1 x and v = cot-1 x

Differentiating with respect to x, we get

\(\rm \frac{du}{dx} = \frac{d(\tan^{-1} x)}{dx} = \frac{1}{1+x^2}\)

\(\rm \frac{dv}{dx} = \frac{d(\cot^{-1} x)}{dx} = \frac{-1}{1+x^2}\)

Now,

\(\rm \frac{d\tan^{-1} x}{d\cot^{-1} x} = \frac{du}{dv}\\=\frac{\frac{du}{dx}}{\frac{dv}{dx}}\\=\frac{\frac{1}{1+x^2}}{\frac{-1}{1+x^2}}=-1\)