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This is the second of our many quizzes on Chain rule. Click on this to access more such competitive quizzes on Calculus.

Chain rule Quiz 2

mathematics

   What will be the numerator of the value of $\dfrac{d}{dx}\left( \ln \left( 3{{x}^{2}} \right) \right)$.

  Apply the formula, $\dfrac{d}{dx}\left( \ln (u) \right)=\dfrac{1}{u}\dfrac{d}{dx}(u)$

  $ \dfrac{d}{dx}\left( \ln \left( 3{{x}^{2}} \right) \right)=\dfrac{1}{3{{x}^{2}}}\dfrac{d}{dx}\left( 3{{x}^{2}} \right) \\  \Rightarrow \dfrac{d}{dx}\left( \ln \left( 3{{x}^{2}} \right) \right)=\dfrac{6x}{3{{x}^{2}}}=\dfrac{2}{x} \\  \text{So, the numerator of the value of} \dfrac{d}{dx}\left( \ln \left( 3{{x}^{2}} \right) \right) \text{is 2.}$

 What is the coefficient of $x$ in derivative of ${{\left( 3{{x}^{2}}+2 \right)}^{2}}$.

 Apply the formula, $\dfrac{d}{dx}\left( u{{\left( x \right)}^{n}} \right)=nu{{\left( x \right)}^{n-1}}\dfrac{d}{dx}\left( u\left( x \right) \right)$

  $ \text{Given:} {{\left( 3{{x}^{2}}+2 \right)}^{2}}, \text{so} \\ \dfrac{d}{dx}\left( {{\left( 3{{x}^{2}}+2 \right)}^{2}} \right)=2\left( 3{{x}^{2}}+2 \right)\dfrac{d}{dx}\left( 3{{x}^{2}}+2 \right) \\  \Rightarrow \dfrac{d}{dx}\left( {{\left( 3{{x}^{2}}+2 \right)}^{2}} \right)=\left( 6{{x}^{2}}+4 \right)\left( 6x \right)=36{{x}^{3}}+24x \\ \text{So, the coefficient of x in derivative of} {{\left( 3{{x}^{2}}+2 \right)}^{2}} \text{is 24.}$

  What is the value of  $\dfrac{d}{dx}\left( {{\left( 2x-9\sqrt{x} \right)}^{2}} \right)$.

 $\dfrac{d}{dx}\left( {{\left( 2x-9\sqrt{x} \right)}^{2}} \right)=2\left( 2x-9\sqrt{x} \right)\dfrac{d}{dx}\left( 2x-9\sqrt{x} \right) \\  \Rightarrow \dfrac{d}{dx}\left( {{\left( 2x-9\sqrt{x} \right)}^{2}} \right)=2\left( 2x-9\sqrt{x} \right)\left( 2-\dfrac{9}{2\sqrt{x}} \right) \\  \therefore \dfrac{d}{dx}\left( {{\left( 2x-9\sqrt{x} \right)}^{2}} \right)=8x-54\sqrt{x}+81 $

 What is the value of  $\dfrac{d}{dx}{{\left( 3x-\sqrt{x} \right)}^{2}}$ when  $x=1$.

  Apply the formula, $\dfrac{d}{dx}\left( u{{\left( x \right)}^{n}} \right)=nu{{\left( x \right)}^{n-1}}\dfrac{d}{dx}\left( u\left( x \right) \right)$. Then substitute $x=1$.

 $ \dfrac{d}{dx}\left( {{\left( 3x-\sqrt{x} \right)}^{2}} \right)=2\left( 3x-\sqrt{x} \right)\dfrac{d}{dx}\left( 3x-\sqrt{x} \right) \\  \Rightarrow \dfrac{d}{dx}\left( {{\left( 3x-\sqrt{x} \right)}^{2}} \right)=2\left( 3x-\sqrt{x} \right)\left( 3-\dfrac{1}{2\sqrt{x}} \right) \\  \therefore \dfrac{d}{dx}\left( {{\left( 3x-\sqrt{x} \right)}^{2}} \right)=18x-9\sqrt{x}+1 \\ <br>\text{Substitute} x=1, \text{we get} \\  \therefore \dfrac{d}{dx}{{\left( {{\left( 3x-\sqrt{x} \right)}^{2}} \right)}_{x=1}}=10 $

  What is the difference between the derivative of $\dfrac{d}{dx}\left( {{\left( \cos x+5 \right)}^{2}}+{{\sin }^{2}}2x \right)$ at $x=\dfrac{\pi }{4}$ and  $x=\dfrac{\pi }{2}$.

 Apply the formulas, $\dfrac{d}{dx}\left( u{{\left( x \right)}^{n}} \right)=nu{{\left( x \right)}^{n-1}}\dfrac{d}{dx}\left( u\left( x \right) \right),\dfrac{d}{dx}\left( \sin x \right)=\cos x,\dfrac{d}{dx}\left( \cos x \right)=-\sin x$. Then substitute $x=\dfrac{\pi }{4}$ and $x=\dfrac{\pi }{2}$. Now get their differences.

$ \dfrac{d}{dx}\left( {{\left( \cos x+5 \right)}^{2}}+{{\sin }^{2}}2x \right)=\dfrac{d}{dx}\left( {{\left( \cos x+5 \right)}^{2}} \right)+\dfrac{d}{dx}\left( {{\sin }^{2}}2x \right) \\  \Rightarrow \dfrac{d}{dx}\left( {{\left( \cos x+5 \right)}^{2}}+{{\sin }^{2}}2x \right)=2\left( \cos x+5 \right)\dfrac{d}{dx}\left( \cos x+5 \right)+2\sin 2x\dfrac{d}{dx}\left( \sin 2x \right) \\  \Rightarrow \dfrac{d}{dx}\left( {{\left( \cos x+5 \right)}^{2}}+{{\sin }^{2}}2x \right)=2\left( \cos x+5 \right)\left( -\sin x \right)+2\sin 2x\left( 2\cos 2x \right) \\  \therefore \dfrac{d}{dx}\left( {{\left( \cos x+5 \right)}^{2}}+{{\sin }^{2}}2x \right)=4\sin 2x\cos 2x-2\sin x\left( \cos x+5 \right) \\ \text{Substitute} x=\dfrac{\pi }{4}\\ \text{we get}\\  \dfrac{d}{dx}{{\left( {{\left( \cos x+5 \right)}^{2}}+{{\sin }^{2}}2x \right)}_{x=\dfrac{\pi }{4}}}=4\sin 2\left( \dfrac{\pi }{4} \right)\cos 2\left( \dfrac{\pi }{4} \right)-2\sin \left( \dfrac{\pi }{4} \right)\left( \cos \left( \dfrac{\pi }{4} \right)+5 \right) \\  \Rightarrow \dfrac{d}{dx}{{\left( {{\left( \cos x+5 \right)}^{2}}+{{\sin }^{2}}2x \right)}_{x=\dfrac{\pi }{4}}}=-2\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{\sqrt{2}}+5 \right)=-1-5\sqrt{2}........(i) \\  \text{Substitute} x=\dfrac{\pi }{2},\\ \text{we get} \\  \dfrac{d}{dx}{{\left( {{\left( \cos x+5 \right)}^{2}}+{{\sin }^{2}}2x \right)}_{x=\dfrac{\pi }{2}}}=4\sin 2\left( \dfrac{\pi }{2} \right)\cos 2\left( \dfrac{\pi }{2} \right)-2\sin \left( \dfrac{\pi }{2} \right)\left( \cos \left( \dfrac{\pi }{2} \right)+5 \right) \\  \Rightarrow \dfrac{d}{dx}{{\left( {{\left( \cos x+5 \right)}^{2}}+{{\sin }^{2}}2x \right)}_{x=\dfrac{\pi }{2}}}=0-2\left( 1 \right)\left( 0+5 \right)=0-10..........(ii) \\  \text{Required difference is}-1-5\sqrt{2}+10=9-5\sqrt{2} $