Question

Differentiate.

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

Short answer

Derivative of the function is \(\frac{1}{{2\sqrt x {{\left( {\sqrt x + 1} \right)}^2}}}\).

Step-by-step solution

Step 1:Find the derivative of y using the Quotient rule

The equation for the Quotient ruleis \(\left( {\frac{f}{g}} \right)' = \frac{{gf' - fg'}}{{{{\left( g \right)}^2}}}\)

Apply Quotient rule for the function \(y = \frac{{\sqrt x }}{{\sqrt x + 1}}\).

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

Simplify the obtained condition

Simplify the obtained derivative as:

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

So, \(y' = \frac{1}{{2\sqrt x {{\left( {\sqrt x + 1} \right)}^2}}}\)is the final answer of this derivative function.