Question

Prove the formula for \(\frac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right)\) by the same method for \(\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right)\).

Short answer

The formula for \(\frac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right)\) is proved by the same method for \(\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right)\).

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\).

In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

The derivative of the function

Let \(y = {\cos ^{ - 1}}x\). This implies, \(\cos y = x\). Differentiate both the sides of the above equation with respect to \(x\) as follows:

\(\begin{aligned}\frac{d}{{dx}}\left( {\cos y} \right) & = \frac{d}{{dx}}\left( x \right)\\ - \sin y\frac{{dy}}{{dx}} & = 1\\\frac{{dy}}{{dx}} & = - \frac{1}{{\sin y}}\\\frac{{dy}}{{dx}} & = - \frac{1}{{\sqrt {1 - {{\cos }^2}y} }}\\\frac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) &= - \frac{1}{{\sqrt {1 - {x^2}} }}\end{aligned}\)

Hence, the formula for \(\frac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right)\) is proved.