Question

Find the derivative of the function. Simplify where possible.

78. \(y = \arctan \sqrt {\frac{{1 - x}}{{1 + x}}} \)

Short answer

The derivative of the function is \(\frac{{ - 1}}{{2\sqrt {1 - {x^2}} }}\).

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 given function

Differentiate \(y = \arctan \sqrt {\frac{{1 - x}}{{1 + x}}} \) with respect to \(x\) as follows:

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {\arctan \sqrt {\frac{{1 - x}}{{1 + x}}} } \right)\\ &= \frac{1}{{1 + {{\left( {\sqrt {\frac{{1 - x}}{{1 + x}}} } \right)}^2}}} \cdot \frac{d}{{dx}}\left( {\sqrt {\frac{{1 - x}}{{1 + x}}} } \right)\\ &= \frac{1}{{1 + \frac{{1 - x}}{{1 + x}}}} \cdot \left( {\frac{1}{{2\sqrt {\frac{{1 - x}}{{1 + x}}} }}} \right) \cdot \frac{d}{{dx}}\left( {\frac{{1 - x}}{{1 + x}}} \right)\\ &= \frac{1}{{\frac{{1 + x + 1 - x}}{{1 + x}}}} \cdot \left( {\frac{1}{2}\sqrt {\frac{{1 + x}}{{1 - x}}} } \right) \cdot \left( {\frac{{\left( {1 + x} \right)\frac{d}{{dx}}\left( {1 - x} \right) - \left( {1 - x} \right)\frac{d}{{dx}}\left( {1 + x} \right)}}{{{{\left( {1 + x} \right)}^2}}}} \right)\end{aligned}\)

Solve the above equation further as follows:

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{{1 + x}}{2} \cdot \left( {\frac{1}{2}\sqrt {\frac{{1 + x}}{{1 - x}}} } \right) \cdot \left( {\frac{{\left( {1 + x} \right)\left( { - 1} \right) - \left( {1 - x} \right)\left( 1 \right)}}{{{{\left( {1 + x} \right)}^2}}}} \right)\\ &= \frac{{1 + x}}{2} \cdot \left( {\frac{1}{2}\sqrt {\frac{{1 + x}}{{1 - x}}} } \right) \cdot \left( {\frac{{ - 2}}{{{{\left( {1 + x} \right)}^2}}}} \right)\\ &= - \frac{{1 + x}}{2} \cdot \left( {\sqrt {\frac{{1 + x}}{{1 - x}}} } \right) \cdot \left( {\frac{1}{{{{\left( {1 + x} \right)}^2}}}} \right)\\ &= \frac{{ - 1}}{{2\sqrt {1 - x} \sqrt {1 + x} }}\\ &= \frac{{ - 1}}{{2\sqrt {1 - {x^2}} }}\end{aligned}\)

Hence, the derivative of \(y\) is \(\frac{{ - 1}}{{2\sqrt {1 - {x^2}} }}\).