# Find \frac{dy}{dx} in simplest factored form 1) \frac{\sqrt{2x + 1}}{x^2 - 1} \\ 2)...

## Question:

Find {eq}\frac{dy}{dx} {/eq} in simplest factored form

{eq}1) \frac{\sqrt{2x + 1}}{x^2 - 1} \\ 2) \frac{\ln(x^2 + 5)}{x^2 + 5} \\ 3) e^{x + y} {/eq}

## Chain Rule of Differentiation:

We have to find the derivative of three given function. In first two function we will use quotient rule differentiation and chain rule of differentiation. In third function we will use chain rule of differentiation to get the desired results.

1.) Use the quotient rule of differentiation and we have \begin{align*} y &= \frac{{\sqrt {2x + 1} }}{{{x^2} - 1}}\\ \frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {\frac{{\sqrt {2x + 1} }}{{{x^2} - 1}}} \right)\\ \frac{{dy}}{{dx}} &= \frac{{\left( {{x^2} - 1} \right) \cdot \frac{d}{{dx}}\left( {\sqrt {2x + 1} } \right) - \frac{d}{{dx}}\left( {{x^2} - 1} \right) \cdot \sqrt {2x + 1} }}{{{{\left( {{x^2} - 1} \right)}^2}}}\\ \frac{{dy}}{{dx}} &= \frac{{\left( {{x^2} - 1} \right) \cdot \frac{1}{{2\sqrt {2x + 1} }} \cdot \frac{d}{{dx}}\left( {2x + 1} \right) - 2x \cdot \sqrt {2x + 1} }}{{{{\left( {{x^2} - 1} \right)}^2}}}\\ \frac{{dy}}{{dx}} &= \frac{{\left( {{x^2} - 1} \right) \cdot \frac{1}{{2\sqrt {2x + 1} }} \cdot 2 - 2x \cdot \sqrt {2x + 1} }}{{{{\left( {{x^2} - 1} \right)}^2}}}\\ \frac{{dy}}{{dx}} &= \frac{{\frac{{\left( {{x^2} - 1} \right) - 2x\left( {2x + 1} \right)}}{{\sqrt {2x + 1} }}}}{{{{\left( {{x^2} - 1} \right)}^2}}}\\ \frac{{dy}}{{dx}} &= \frac{{\left( {{x^2} - 1} \right) - 4{x^2} - 2x}}{{{{\left( {{x^2} - 1} \right)}^2} \cdot \sqrt {2x + 1} }}\\ \frac{{dy}}{{dx}} &= \frac{{ - 3{x^2} - 2x - 1}}{{{{\left( {{x^2} - 1} \right)}^2} \cdot \sqrt {2x + 1} }}. \end{align*}

2.) Use the quotient rule of differentiation and we have \begin{align*} y &= \frac{{\ln ({x^2} + 5)}}{{{x^2} + 5}}\\ \frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left[ {\frac{{\ln \left( {{x^2} + 5} \right)}}{{{x^2} + 5}}} \right]\\ \frac{{dy}}{{dx}} &= \frac{{\left( {{x^2} + 5} \right) \cdot \frac{d}{{dx}}\left[ {\ln \left( {{x^2} + 5} \right)} \right] - \frac{d}{{dx}}\left( {{x^2} + 5} \right) \cdot \ln \left( {{x^2} + 5} \right)}}{{{{\left( {{x^2} + 5} \right)}^2}}}\\ \frac{{dy}}{{dx}} &= \frac{{\left( {{x^2} + 5} \right) \cdot \frac{1}{{\left( {{x^2} + 5} \right)}} \cdot \frac{d}{{dx}}\left[ {{x^2} + 5} \right] - 2x \cdot \ln \left( {{x^2} + 5} \right)}}{{{{\left( {{x^2} + 5} \right)}^2}}}\\ \frac{{dy}}{{dx}} &= \frac{{2x - 2x \cdot \ln \left( {{x^2} + 5} \right)}}{{{{\left( {{x^2} + 5} \right)}^2}}}\\ \frac{{dy}}{{dx}} &= \frac{{2x\left[ {1 - \ln \left( {{x^2} + 5} \right)} \right]}}{{{{\left( {{x^2} + 5} \right)}^2}}}. \end{align*}

3.) Use chain rule of differentiation and we have \begin{align*} y &= {e^{x + y}}\\ \frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{e^{x + y}}} \right)\\ \frac{{dy}}{{dx}} &= {e^{x + y}} \cdot \frac{d}{{dx}}\left( {x + y} \right)\\ \frac{{dy}}{{dx}} &= {e^{x + y}} \cdot \left( {1 + \frac{{dy}}{{dx}}} \right)\\ \frac{{dy}}{{dx}}\left( {1 - {e^{x + y}}} \right) &= {e^{x + y}}\\ \frac{{dy}}{{dx}} &= \frac{{{e^{x + y}}}}{{1 - {e^{x + y}}}}. \end{align*}