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This is the second of our many quizzes on Derivative definition and notation. Click on this to access more such competitive quizzes on Calculus.

Derivative definition and notation Quiz 2

mathematics

 Use the standard definition of Leibniz notation that is derivation of a function is the rate of change in ‘y’ with respect to rate of change in ‘x' and use the representation of derivative and check the options for the correct statements.

  Given options a and b states Leibniz rule in two different ways. Both the options are correct, so option c is the correct answer.

 If $f\ \left( x \right)=\cos 100x$and ${f}'\left( x \right)=\underset{\Delta x\to 0}{\mathop{\lim }}\,a\dfrac{\left( \cos 100\Delta x-1 \right)}{\Delta x}-\underset{x\to 0}{\mathop{\lim }}\,b\dfrac{\sin 100\Delta x}{\Delta x}$ then find the value of ${{a}^{2}}+{{b}^{2}}$?

 From the derivation notation solve $f\left( x \right)$ and then in the process of solving we can find the similar step that was asked in the question. After that, solve the required equation to get the answer.

  $\text{Finding the derivative using derivative notation}\text{.} \\ \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{f\left( x+\Delta x \right)-f\left( x \right)}{\Delta x} \\ f\left( x+\Delta x \right)-f\left( x \right)=\cos 100\left( x+\Delta x \right)-\cos 100x \\ \Rightarrow \cos 100x\cos 100\Delta x-\sin 100x\sin 100\Delta x-\cos 100x \\ \Rightarrow \cos 100x\left( \cos 100\Delta x-1 \right)-\sin 100x\sin 100\Delta x \\ \text{Therefore} \\ {{f}^{'}}\left( x \right)=\underset{\Delta x\to 0}{\mathop{\lim }}\,\cos 100x\dfrac{\left( \cos 100\Delta x-1 \right)}{\Delta x}-\underset{x\to 0}{\mathop{\lim }}\,\sin 100x\dfrac{\sin 100\Delta x}{\Delta x} \\ \text{Therefore a and b are cos100x and sin100x respectively}\text{.} \\ \Rightarrow {{\text{a}}^{2}}+{{b}^{2}}=1 $

 If $f\left( x \right)=\dfrac{254}{\sin x}$and ${{f}^{'}}\left( x \right)=\left( a\sin x\dfrac{\left( \cos \Delta x-1 \right)}{\Delta x}-b\dfrac{\sin \Delta x}{\Delta x}\cos x \right)\left( \dfrac{1}{\sin \left( x+\Delta x \right)\sin x} \right)$ then find out the value of $a+b$? 

$\text{Finding the derivative using derivative notation}\text{.} \\ \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{f\left( x+\Delta x \right)-f\left( x \right)}{\Delta x} \\ f\left( x+\Delta x \right)-f\left( x \right)=\dfrac{254}{\sin \left( x+\Delta x \right)}-\dfrac{254}{\sin \left( x \right)} \\ \Rightarrow \dfrac{254\sin x-254\sin x\cos \Delta x-254\sin \Delta x\cos x}{\sin \left( x+\Delta x \right)\sin x} \\ \Rightarrow \dfrac{-254\sin x\left( \cos \Delta x-1 \right)-254\sin \Delta x\cos x}{\sin \left( x+\Delta x \right)\sin x} \\ \text{Therefore} \\ {{f}^{'}}\left( x \right)=\left( -254\sin x\dfrac{\left( \cos \Delta x-1 \right)}{\Delta x}-254\dfrac{\sin \Delta x}{\Delta x}\cos x \right)\left( \dfrac{1}{\sin \left( x+\Delta x \right)\sin x} \right) \\ \text{So, }a+b=0 $

 Solve $\left( {{y}^{3}}{{x}^{2}}+{{y}^{2}} \right)dx+\left( 2{{x}^{3}}{{y}^{2}}-xy \right)dy=0$. 

 Expand the given equation and then apply the formula $d\left( x{{y}^{2}} \right)={{y}^{2}}dx+2xydy$ and $d\left( \dfrac{y}{x} \right)=\dfrac{ydx-xdy}{{{x}^{2}}}$ to get the solution.

 Given, $\Rightarrow \left( {{y}^{3}}{{x}^{2}}+{{y}^{2}} \right)dx+\left( 2{{x}^{3}}{{y}^{2}}-xy \right)dy=0 \\ $ By expanding, $ \\ \Rightarrow {{y}^{3}}{{x}^{2}}dx+2{{x}^{3}}{{y}^{2}}dy+{{y}^{2}}dx-xydy=0 \\ $ Here $a=1$ and $b=2$. Thus, $ \\ \Rightarrow d\left( x{{y}^{2}} \right)={{y}^{2}}dx+2xydy \\ $ This suggests dividing the whole equation by ${{x}^{2}}y$ $ \\ \Rightarrow {{y}^{2}}dx+2xydy+\dfrac{ydx-xdy}{{{x}^{2}}}=0 \\ $ Thus, $x{{y}^{2}}+\dfrac{y}{x}=\text{C}$

 If $y=99\tan \left( \cos \left( 99\sin x \right) \right)$, then find $\dfrac{dy}{dx}$? 

 By applying the chain rule we can simply solve the problem. After applying the chain rule, solve the respective equation to get the solution.

 $\text{For }y=99\tan \left( \cos \left( 99\sin x \right) \right) \\ \text{Let us assume }y=\tan u,u=\cos v\text{ and }v=99\sin x \\ \text{By applying the chain rule } \\ \Rightarrow \dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dv}.\dfrac{dv}{dx} \\ \Rightarrow \dfrac{d}{du}\left( 99\tan u \right).\dfrac{d}{dv}\left( \cos v \right)\dfrac{d}{dx}\left( 99\sin x \right) \\ \text{By applying trigonometric formulas} \\ \Rightarrow -9801\sec ^2\left(\cos \left(99\sin \left(x\right)\right)\right)\sin \left(99\sin \left(x\right)\right)\cos \left(x\right) $