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This is one of our many quizzes on Power rule. Click on this to access more such competitive quizzes on Calculus.

Power rule Quiz 2

mathematics

  Evaluate $\underset{v\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{v}}+\sin v-1}{\log \left( 1+v \right)}$

  The given limit is in indeterminate form i.e.  $\dfrac{0}{0}$ so, we need to apply L’ Hospital’s Rule i.e., differentiate both the numerator and the denominator and then apply the limit.  

$\underset{v\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{v}}+\sin v-1}{\log \left( 1+v \right)}\text{ }\left[ \text{Form }\dfrac{0}{0} \right] \\ =\underset{v\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{v}}+\cos v}{\dfrac{1}{\left( 1+v \right)}}=\underset{v\to 0}{\mathop{\lim }}\,\left( 1+v \right)\left( {{e}^{v}}+\cos v \right) \\ =2  $

 Determine the nature of  $\phi \left( b \right)=\dfrac{b}{{{e}^{b}}}$ (increasing/decreasing) at the points $b=-4,\text{ }b=0$. Use ${{e}^{4}}=54.598$.

 In order to find the nature of the function we need to find it’s derivative which can be found out using quotient rule and power rule. Check if $\phi '\left( b \right)>0$, if so then the function is increasing.

 $ \phi \left( b \right)=\dfrac{b}{{{e}^{b}}} \\ \Rightarrow \phi '\left( b \right)=\dfrac{\left( 1 \right){{e}^{b}}-b\left( {{e}^{b}} \right)}{{{\left( {{e}^{b}} \right)}^{2}}} \\ \Rightarrow \phi '\left( b \right)=\dfrac{{{e}^{b}}-b{{e}^{b}}}{{{\left( {{e}^{b}} \right)}^{2}}}=\dfrac{1-b}{{{e}^{b}}} \\ \Rightarrow \phi '\left( -4 \right)=\dfrac{1-\left( -4 \right)}{{{e}^{-4}}}=5{{e}^{4}}=5\times 54.598\approx 273>0\text{ }\therefore \text{Increasing} \\ \Rightarrow \phi '\left( 0 \right)=\dfrac{1-0}{{{e}^{0}}}=1>0\text{ }\therefore \text{Increasing}$

 Find  $p'$ with the help of implicit differentiation for $7{{p}^{2}}+\sin \left( 3q \right)=12-{{p}^{4}}$.

 Differentiate both the sides using power rule $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$and $\dfrac{d}{dx}\sin x=\cos x$.

 $ 7{{p}^{2}}+\sin \left( 3q \right)=12-{{p}^{4}} \\ \text{Differentiating both sides w}\text{.r}\text{.t q} \\ \Rightarrow \text{14pp }\!\!'\!\!\text{ +3cos}\left( 3q \right)=-4{{p}^{3}}p' \\ \Rightarrow \left( 14p+4{{p}^{3}} \right)p'=-3\cos \left( 3q \right) \\ \Rightarrow p'=\dfrac{-3\cos \left( 3q \right)}{\left( 14p+4{{p}^{3}} \right)} $

 If  $\phi \left( v \right)={{v}^{3}}-6{{v}^{2}}+24v+4$ then find its maximum value.

 In order to find maximum value, we need to find $\phi '\left( v \right)$ and equate it with zero.

 $ \phi \left( v \right)={{v}^{3}}-6{{v}^{2}}+24v+4 \\ \Rightarrow \phi '\left( v \right)=3{{v}^{2}}-12v+24 \\ \phi '\left( v \right)=0\left[ \text{for maximum value} \right] \\ \Rightarrow 3{{v}^{2}}-12v+24=0 \\ \Rightarrow {{v}^{2}}-4v+8=0 \\ \Rightarrow v=2\pm 2i \\ \because \text{The equation }\phi '\left( b \right)=0\text{ has no real roots}\text{.} \\ \text{So }\phi \left( b \right)\text{ has neither a  maximum nor  a minimum}\text{. } $

 Determine the value of $c$ which satisfy the Rolle’s theorem for $\varphi \left( k \right)={{k}^{2}}-2k-8$  for $\left[ -1,3 \right]$.

 Rolle’s theorem states that if a function $f$  is continuous on $\left[ a,b \right]$and differentiable $\left( a,b \right)$ such that $f\left( a \right)=f\left( b \right)$ then$f'\left( x \right)=0$ for some $x$ with $a\le x\le b$.

 $  \varphi \left( k \right)={{k}^{2}}-2k-8 \\ \text{The function is polynomial } \\ \therefore \text{ It is continuous on }\left[ -1,3 \right]\text{ and differentiable on }\left( -1,3 \right). \\ \text{Again }\varphi \left( -1 \right)=\varphi \left( 3 \right)=-5. \\ \varphi '\left( k \right)=2k-2 \\ \Rightarrow \varphi '\left( c \right)=0\left[ \text{According to Rolle }\!\!'\!\!\text{ s theorem} \right] \\ \Rightarrow 2c-2=0 \\ \Rightarrow c=1 $