# Additional Math – Differentiation – Quotient Rule (Challenging)

Differentiate $latex \displaystyle\ y=\frac{{{{x}^{2}}\sqrt{{x+1}}}}{{x-1}}$ with respect to x.

Simplify the Numerator (otherwise you need to use both quotient rule for the expression and product rule for the numerator).

$latex \displaystyle y=\frac{{{{{\left( {{{x}^{4}}} \right)}}^{{\frac{1}{2}}}}{{{\left( {x+1} \right)}}^{{\frac{1}{2}}}}}}{{x-1}}$

$latex \displaystyle y=\frac{{{{{\left[ {{{x}^{4}}\left( {x+1} \right)} \right]}}^{{\frac{1}{2}}}}}}{{x-1}}$

$latex \displaystyle y=\frac{{{{{\left[ {{{x}^{5}}+{{x}^{4}}} \right]}}^{{\frac{1}{2}}}}}}{{x-1}}$

Use the quotient rule formula $latex \displaystyle \frac{{dy}}{{dx}}=\frac{{v.\frac{{du}}{{dx}}-u.\frac{{dv}}{{dx}}}}{{{{v}^{2}}}}&fg=ff0000$

$latex \displaystyle u={{\left( {{{x}^{5}}+{{x}^{4}}} \right)}^{{\frac{1}{2}}}}&fg=ff0000$

$latex \displaystyle \frac{{du}}{{dx}}=\frac{1}{2}{{\left( {{{x}^{5}}+{{x}^{4}}} \right)}^{{-\frac{1}{2}}}}\left( {5{{x}^{4}}+4{{x}^{3}}} \right)&fg=ff0000$

$latex \displaystyle v=x-1&fg=ff0000$

$latex \displaystyle \frac{{dv}}{{dx}}=1&fg=ff0000$

$latex \displaystyle \frac{{dy}}{{dx}}=\frac{{(x-1)\left( {\frac{1}{2}} \right){{{\left( {{{x}^{5}}+{{x}^{4}}} \right)}}^{{-\frac{1}{2}}}}\left( {5{{x}^{4}}+4{{x}^{3}}} \right)-{{{\left( {{{x}^{5}}+{{x}^{4}}} \right)}}^{{\frac{1}{2}}}}\left( 1 \right)}}{{{{{\left( {x-1} \right)}}^{2}}}}$

Factorize $latex \displaystyle {{{{\left( {{{x}^{5}}+{{x}^{4}}} \right)}}^{{-\frac{1}{2}}}}}&fg=ff0000$ from the numerator and bring it down to the denominator.

$latex \displaystyle \frac{{dy}}{{dx}}=\frac{{{{{\left( {{{x}^{5}}+{{x}^{4}}} \right)}}^{{-\frac{1}{2}}}}\left[ {(x-1)\left( {\frac{1}{2}} \right)\left( {5{{x}^{4}}+4{{x}^{3}}} \right)-\left( {{{x}^{5}}+{{x}^{4}}} \right)} \right]}}{{{{{\left( {x-1} \right)}}^{2}}}}$

$latex \displaystyle \frac{{dy}}{{dx}}=\frac{{\left( {x-1} \right)\left( {\frac{1}{2}} \right)\left( {5{{x}^{4}}+4{{x}^{3}}} \right)-\left( {{{x}^{5}}+{{x}^{4}}} \right)}}{{{{{\left( {x-1} \right)}}^{2}}{{{\left( {{{x}^{5}}+{{x}^{4}}} \right)}}^{{\frac{1}{2}}}}}}$

Simplify the numerator once more.

$latex \displaystyle \frac{{dy}}{{dx}}=\frac{{\frac{5}{2}{{x}^{5}}+2{{x}^{4}}-\frac{5}{2}{{x}^{4}}-2{{x}^{3}}-{{x}^{5}}-{{x}^{4}}}}{{{{{\left( {x-1} \right)}}^{2}}{{{\left( {{{x}^{5}}+{{x}^{4}}} \right)}}^{{\frac{1}{2}}}}}}$

$latex \displaystyle \frac{{dy}}{{dx}}=\frac{{\frac{3}{2}{{x}^{5}}-\frac{3}{2}{{x}^{4}}-2{{x}^{3}}}}{{{{{\left( {x-1} \right)}}^{2}}{{{\left( {{{x}^{5}}+{{x}^{4}}} \right)}}^{{\frac{1}{2}}}}}}$

$latex \displaystyle \frac{{dy}}{{dx}}=\frac{{\frac{1}{2}\left( {3{{x}^{5}}-3{{x}^{4}}-4{{x}^{3}}} \right)}}{{{{{\left( {x-1} \right)}}^{2}}{{{\left( {{{x}^{5}}+{{x}^{4}}} \right)}}^{{\frac{1}{2}}}}}}$

$latex \displaystyle \frac{{dy}}{{dx}}=\frac{{\left( {3{{x}^{5}}-3{{x}^{4}}-4{{x}^{3}}} \right)}}{{2{{{\left( {x-1} \right)}}^{2}}{{{\left( {{{x}^{5}}+{{x}^{4}}} \right)}}^{{\frac{1}{2}}}}}}$

Additional Math A-Math , Secondary 1 and Secondary 2 Math Tuition Small Group Tuition at Woodlands.

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