Calculus : Differentiation (Chain Rule, Product Rule and Quotient Rule)

Differentiate the following with respect to $ \displaystyle x$.

(i)  $ \displaystyle (2−x^2) \ln(3x+1)$

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(i) $ \displaystyle \ \ \ \ \frac{d}{{dx}}\left[ {(2-{{x}^{2}})\ln (3x+1)} \right]$

$ \displaystyle =(2-{{x}^{2}})\frac{d}{{dx}}\ln (3x+1)+\ln (3x+1)\frac{d}{{dx}}(2-{{x}^{2}})$

$ \displaystyle =\frac{{2-{{x}^{2}}}}{{3x+1}}\frac{d}{{dx}}\left( {3x+1} \right)-2x\ln (3x+1)$

$ \displaystyle =\frac{{3\left( {2-{{x}^{2}}} \right)}}{{3x+1}}-2x\ln (3x+1)$


(ii)  $ \displaystyle {\frac{{4-\tan 2x}}{{5x}}}$

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(ii) $ \displaystyle \ \ \ \ \frac{d}{{dx}}\left[ {\frac{{4-\tan 2x}}{{5x}}} \right]$

$ \displaystyle =\frac{{5x\frac{d}{{dx}}\left( {4-\tan 2x} \right)-\left( {4-\tan 2x} \right)\frac{d}{{dx}}(5x)}}{{25{{x}^{2}}}}$

$ \displaystyle =\frac{{5x(-{{{\sec }}^{2}}2x)\frac{d}{{dx}}(2x)-5\left( {4-\tan 2x} \right)}}{{25{{x}^{2}}}}$

$ \displaystyle =\frac{{-10x{{{\sec }}^{2}}2x+5\tan 2x-20}}{{25{{x}^{2}}}}$

(iii)  Given that a curve has equation $ \displaystyle y=\frac{1}{x}+2\sqrt{x}$ where $\ x>0$, Find $ \displaystyle \frac{{dy}}{{dx}}$ and $ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$.
Hence or otherwise find the coordinates and nature of the stationary point of the curve.

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(iii)   $ \displaystyle y=\frac{1}{x}+2\sqrt{x},x>0$

$ \displaystyle \frac{{dy}}{{dx}}=-\frac{1}{{{{x}^{2}}}}+\frac{1}{{\sqrt{x}}}$

$ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\frac{2}{{{{x}^{3}}}}-\frac{1}{{2x\sqrt{x}}}$

At the stationary point, $ \displaystyle \frac{{dy}}{{dx}}=0$.

$ \displaystyle \therefore -\frac{1}{{{{x}^{2}}}}+\frac{1}{{\sqrt{x}}}=0$

$ \displaystyle \therefore \frac{1}{{\sqrt{x}}}=\frac{1}{{{{x}^{2}}}}$

$ \displaystyle \therefore x\sqrt{x}=1$

$ \displaystyle \therefore x=1$

When $ \displaystyle x=1$, $ \displaystyle y=\frac{1}{1}+2\sqrt{1}=3$.

Therefore, the stationary point is $ \displaystyle (1,3)$.

$ \displaystyle {{\left. {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right|}_{{(1,3)}}}=\frac{2}{{{{1}^{2}}}}-\frac{1}{{2(1)\sqrt{1}}}=\frac{3}{2}>0$

Therefore, the stationary point (1,3) is a minimum turning point.