Question

Find the derivative of the function:

23. \(y = \sqrt {\frac{x}{{x + 1}}} \)

Short answer

The derivative of y is \(\frac{1}{2\sqrt{x}{{\left( x+1 \right)}^{{3}/{2}\;}}}\).

Step-by-step solution

The quotient rule of differentiation

If a function\(h\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\)and\(f\),\(g\)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 \(y\) with respect to \(x\) as follows:

\begin{aligned}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left[ {{{\left( {\frac{x}{{x + 1}}} \right)}^{{1 \mathord{\left/{\vphantom {1 2}} \right.} 2}}}} \right] \\ &= \frac{1}{2}{\left( {\frac{x}{{x + 1}}} \right)^{{{ - 1} \mathord{\left/{\vphantom {{ - 1} 2}} \right.} 2}}}\frac{d}{{dx}}\left( {\frac{x}{{x + 1}}} \right) \\ &= \frac{1}{2}{\left( {\frac{x}{{x + 1}}} \right)^{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} 2}} \right.} 2}}}\left[ {\frac{{\left( {x + 1} \right)\frac{d}{{dx}}\left( x \right) - x\frac{d}{{dx}}\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}} \right) \\ &= \frac{1}{2}{\left( {\frac{x}{{x + 1}}} \right)^{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} 2}} \right.} 2}}}\left[ {\frac{{\left( {x + 1} \right)\left( 1 \right) - x\left( 1 \right)}}{{{{\left( {x + 1} \right)}^2}}}} \right] \\ \end{aligned}

Solve the above equation further,

\begin{aligned}\frac{{dy}}{{dx}} &= \frac{1}{2}{\left( {\frac{x}{{x + 1}}} \right)^{{{ - 1} \mathord{\left/{\vphantom {{ - 1} 2}} \right.} 2}}}\left[ {\frac{{x + 1 - x}}{{{{\left( {x + 1} \right)}^2}}}} \right] \\ &= \frac{1}{2}{\left( {\frac{x}{{x + 1}}} \right)^{{{ - 1} \mathord{\left/{\vphantom {{ - 1} 2}} \right.} 2}}}\left[ {\frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right] \\ &= \frac{1}{2}{\left( {\frac{{x + 1}}{x}} \right)^{{1 \mathord{\left/{\vphantom {1 2}} \right.} 2}}}\left[ {\frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right] \\ &= \frac{1}{{2\sqrt x {{\left( {x + 1} \right)}^{{3 \mathord{\left/{\vphantom {3 2}} \right.} 2}}}}} \\ \end{aligned}

Hence, the derivative of y \(\frac{1}{2\sqrt{x}{{\left( x+1 \right)}^{{3}/{2}\;}}}\).