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This is one of our many quizzes on Constant rule and constant multiple rule. Click on this to access more such competitive quizzes on Differentiation.

Constant rule and constant multiple rule Quiz 2

mathematics

 Find $y'$, if $y=9{{x}^{5}}+7{{e}^{x}}$.

 Differentiation taking one term at a time, by using the formula $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\text{, }\dfrac{d}{dx}\left( k{{x}^{n}} \right)\text{=}k\dfrac{d}{dx}\left( {{x}^{n}} \right)\text{ and }\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$

  $ y=9{{x}^{5}}+7{{e}^{x}} \\  \Rightarrow \dfrac{dy}{dx}=9\dfrac{d}{dx}\left( {{x}^{5}} \right)+7\dfrac{d}{dx}\left( {{e}^{x}} \right) \\  \Rightarrow \dfrac{dy}{dx}=45{{x}^{4}}+7{{e}^{x}} \\ $

 Find $y'$, if $y=2t\sin \left( {{e}^{t}} \right)\cos \left( {{e}^{t}} \right)$.

 First use the identity $\sin 2x=2\sin x\cos x$, then differentiation using the product rule formula.

  $ y=2t\sin \left( {{e}^{t}} \right)\cos \left( {{e}^{t}} \right) \\  \Rightarrow y=t\sin \left( 2{{e}^{t}} \right)\left[ \because \sin 2x=2\sin x\cos x \right] \\  \Rightarrow \dfrac{dy}{dt}=t\dfrac{d}{dt}\left[ \sin \left( 2{{e}^{t}} \right) \right]+\sin \left( 2{{e}^{t}} \right)\dfrac{d}{dt}\left( t \right) \\  \Rightarrow \dfrac{dy}{dt}=t\cos \left( 2{{e}^{t}} \right)\dfrac{d}{dt}\left( 2{{e}^{t}} \right)+\sin \left( 2{{e}^{t}} \right) \\  \Rightarrow \dfrac{dy}{dt}=2t{{e}^{t}}\cos \left( 2{{e}^{t}} \right)+\sin \left( 2{{e}^{t}} \right) \\ $

 Find $y'-y$, if $y{{e}^{-x}}=\sqrt{\tan \left( {{x}^{2}}+47x+5001 \right)}$

 Take $e^{-1}$ on the RHS and differentiate the equation using the product rule formula.

$ y{{e}^{-x}}=\sqrt{\tan \left( {{x}^{2}}+47x+5001 \right)} \\  \Rightarrow \dfrac{dy}{dx}={{e}^{x}}\dfrac{d}{dx}\left[ {{\tan }^{\frac{1}{2}}}\left( {{x}^{2}}+47x+5001 \right) \right]+{{\tan }^{\frac{1}{2}}}\left( {{x}^{2}}+47x+5001 \right)\dfrac{d}{dx}{{e}^{x}} \\  \Rightarrow \dfrac{dy}{dx}={{e}^{x}}\dfrac{1}{2}\left[ {{\tan }^{-\frac{1}{2}}}\left( {{x}^{2}}+47x+5001 \right) \right]\dfrac{d}{dx}\left( \tan \left( {{x}^{2}}+47x+5001 \right) \right)+{{e}^{x}}{{\tan }^{\frac{1}{2}}}\left( {{x}^{2}}+47x+5001 \right) \\  \Rightarrow \dfrac{dy}{dx}={{e}^{x}}\dfrac{{{\sec }^{2}}\left( {{x}^{2}}+47x+5001 \right)\dfrac{d}{dx}\left( {{x}^{2}}+47x+5001 \right)}{2\sqrt{\tan \left( {{x}^{2}}+47x+5001 \right)}}+{{e}^{x}}\sqrt{\tan \left( {{x}^{2}}+47x+5001 \right)} \\  \Rightarrow \dfrac{dy}{dx}={{e}^{x}}\dfrac{{{\sec }^{2}}\left( {{x}^{2}}+47x+5001 \right)\left( 2x+47 \right)}{2\sqrt{\tan \left( {{x}^{2}}+47x+5001 \right)}}+y \\  \Rightarrow y'-y={{e}^{x}}\dfrac{{{\sec }^{2}}\left( {{x}^{2}}+47x+5001 \right)\left( 2x+47 \right)}{2\sqrt{\tan \left( {{x}^{2}}+47x+5001 \right)}} \\ $

 Find $y'$, if $y=\sin \left( 5x \right)\left( 5{{x}^{2}}+1001.2 \right)\sqrt{\left( 5{{x}^{2}}+1001.2 \right)}$

 Differentiate the given equation using the product rule formula.

 $ y=\sin \left( 5x \right)\left( 5{{x}^{2}}+1001.2 \right)\sqrt{\left( 5{{x}^{2}}+1001.2 \right)}=\sin \left( 5x \right){{\left( 5{{x}^{2}}+1001.2 \right)}^{\frac{3}{2}}} \\  \Rightarrow \dfrac{dy}{dx}=\sin \left( 5x \right)\dfrac{d}{dx}\left[ {{\left( 5{{x}^{2}}+1001.2 \right)}^{\frac{3}{2}}} \right]+{{\left( 5{{x}^{2}}+1001.2 \right)}^{\frac{3}{2}}}\dfrac{d}{dx}\left[ \sin \left( 5x \right) \right] \\  \Rightarrow \dfrac{dy}{dx}=\dfrac{3}{2}\sin \left( 5x \right){{\left( 5{{x}^{2}}+1001.2 \right)}^{\frac{1}{2}}}\dfrac{d}{dx}\left( 5{{x}^{2}}+1001.2 \right)+{{\left( 5{{x}^{2}}+1001.2 \right)}^{\frac{3}{2}}}\cos \left( 5x \right)\dfrac{d}{dx}\left( 5x \right) \\  \Rightarrow \dfrac{dy}{dx}=\dfrac{3}{2}\left( 10x \right)\sin \left( 5x \right){{\left( 5{{x}^{2}}+1001.2 \right)}^{\frac{1}{2}}}+5{{\left( 5{{x}^{2}}+1001.2 \right)}^{\frac{3}{2}}}\cos \left( 5x \right) \\  \Rightarrow \dfrac{dy}{dx}=5{{\left( 5{{x}^{2}}+1001.2 \right)}^{\frac{1}{2}}}\left[ 3x\sin \left( 5x \right)+\left( 5{{x}^{2}}+1001.2 \right)\cos \left( 5x \right) \right] \\ $

 Find $y'$, if $y=\log x\sqrt{3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99}$

 Differentiate the given equation using the product rule formula. To simplify use the formula $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}},\dfrac{d}{dx}\left( k{{x}^{n}} \right)=k\dfrac{d}{dx}\left( {{x}^{n}} \right)\text{, }\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}\text{ and }\dfrac{d}{dx}\left( k \right)=0$

 $ y=\log x\sqrt{3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99} \\  \Rightarrow \dfrac{dy}{dx}=\log x\dfrac{d}{dx}\left[ {{\left( 3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99 \right)}^{\frac{1}{2}}} \right]+\sqrt{3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99}\dfrac{d}{dx}(\log x) \\  \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\log x{{\left( 3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99 \right)}^{\frac{-1}{2}}}\dfrac{d}{dx}\left( 3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99 \right)+\sqrt{3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99}\times \dfrac{1}{x} \\  \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\log x{{\left( 3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99 \right)}^{\frac{-1}{2}}}\left( 3\sqrt{2}{{x}^{\sqrt{2}-1}}+6{{x}^{\frac{1}{2}}} \right)+\dfrac{1}{x}{{\left( 3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99 \right)}^{\frac{1}{2}}} \\  \Rightarrow \dfrac{dy}{dx}={{\left( 3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99 \right)}^{\frac{-1}{2}}}\left[ \dfrac{1}{2}\log x\left( 3\sqrt{2}{{x}^{\sqrt{2}-1}}+6{{x}^{\frac{1}{2}}} \right)+\dfrac{1}{x}\left( 3{{x}^{\sqrt{2}}}+4{{x}^{\frac{3}{2}}}+99 \right) \right] \\ $