Question

Find \(f'\left( x \right)\) and \(f''\left( x \right)\).

34.\(f\left( x \right) = \frac{x}{{1 + \sqrt x }}\)

Short answer

\(f'\left( x \right) = \frac{{2 + \sqrt x }}{{2 + 4\sqrt x + 2x}}\) and \(f''\left( x \right) = - \frac{{3 + 4\sqrt x + x}}{{\sqrt x {{\left( {2 + 4\sqrt x + 2x} \right)}^2}}}\).

Step-by-step solution

The quotient rule of differentiation

If a function \(h\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\) where \(f\) and \(g\) are both differentiable, then the derivative of \(h\left( x \right)\) is as follows:

\(\frac{{dh}}{{dx}} = \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{g\left( x \right) \cdot \frac{d}{{dx}}\left( {f\left( x \right)} \right) - f\left( x \right) \cdot \frac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\)

The derivative of the given function

Differentiate \(f\left( x \right)\) using the quotient rule as follows:

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

Hence, the value of \(f'\left( x \right)\) is\(\frac{{2 + \sqrt x }}{{2 + 4\sqrt x + 2x}}\).

The second derivative of the given function

Differentiate \(f'\left( x \right)\) using the quotient rule as follows:

\(\begin{aligned}f'\left( x \right) & = \frac{{2 + \sqrt x }}{{2 + 4\sqrt x + 2x}}\\f''\left( x \right) & = \frac{{\left( {2 + 4\sqrt x + 2x} \right)\frac{d}{{dx}}\left( {2 + \sqrt x } \right) - \left( {2 + \sqrt x } \right)\frac{d}{{dx}}\left( {2 + 4\sqrt x + 2x} \right)}}{{{{\left( {2 + 4\sqrt x + x} \right)}^2}}}\\ & = \frac{{\left( {2 + 4\sqrt x + 2x} \right)\left( {\frac{1}{{2\sqrt x }}} \right) - \left( {2 + \sqrt x } \right)\left( {\frac{4}{{2\sqrt x }} + 2} \right)}}{{{{\left( {2 + 4\sqrt x + x} \right)}^2}}}\\ & = \frac{{\frac{1}{{\sqrt x }} + 2 + \sqrt x - \frac{4}{{\sqrt x }} - 2 - 2\sqrt x }}{{{{\left( {2 + 4\sqrt x + x} \right)}^2}}}\\ & = \frac{{\frac{{ - 3 - 4\sqrt x - x}}{{\sqrt x }}}}{{{{\left( {2 + 4\sqrt x + x} \right)}^2}}}\\ & = - \frac{{3 + 4\sqrt x + x}}{{\sqrt x {{\left( {2 + 4\sqrt x + 2x} \right)}^2}}}\end{aligned}\)

Hence, the value of \(f''\left( x \right)\) is \( - \frac{{3 + 4\sqrt x + x}}{{\sqrt x {{\left( {2 + 4\sqrt x + 2x} \right)}^2}}}\).