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

Implicit differentiation Quiz 2

mathematics

 What is the implicit derivative of $-3{{e}^{3x-y}}=0$.

 Apply the derivative of all the terms separately with respect to $x$ and simplify.

  $ \dfrac{d}{dx}\left( -3{{e}^{3x-y}} \right)=0 \\ \Rightarrow -3{{e}^{3x-y}}\dfrac{d}{dx}\left( 3x-y \right)=0 \\ \Rightarrow -3{{e}^{3x-y}}(3-y')=0 \\ \Rightarrow y'=3 $

 What is the ratio of the highest coefficient of numerator and denominator in the implicit differential of $4y=2{{x}^{2}}+3x-\sqrt[3]{y}$. 

 Apply the derivative of all the terms separately with respect to $x$, here $\dfrac{dy}{dx}=y'$. Bring all the like terms to one term and get the value of $y'$. Then find the required ratio.

  $ 4\dfrac{d}{dx}\left( y \right)=2\dfrac{d}{dx}\left( {{x}^{2}} \right)+3\dfrac{d}{dx}\left( x \right)-\dfrac{d}{dx}\left( \sqrt[3]{y} \right) \\ \Rightarrow 4y'=4x+3-\dfrac{1}{3{{y}^{\frac{2}{3}}}}y' \\ \Rightarrow 4y'+\dfrac{y'}{3{{y}^{\frac{2}{3}}}}=4x+3 \\ \therefore y'=\dfrac{{{y}^{\frac{2}{3}}}\left( 12x+9 \right)}{12{{y}^{\frac{2}{3}}}+1} \\ \therefore \text{Ratio }=\dfrac{12}{12}=1 $

 Find $y'$ of $x{{e}^{y}}=x-y$ when $x=0$ .

 Apply the derivative of all the terms separately with respect to $x$, here $\dfrac{dy}{dx}=y'$. Use formula derivative of product formula. Bring all the like terms to one term and get the value of $y'$. Then substitute $x=0$ and find the required ratio.

 $ \dfrac{d}{dx}\left( x{{e}^{y}} \right)=\dfrac{d}{dx}\left( x \right)-\dfrac{d}{dx}\left( y \right) \\ \Rightarrow x\dfrac{d}{dx}\left( {{e}^{y}} \right)+{{e}^{y}}=1-y' \\ \Rightarrow x{{e}^{y}}y'+{{e}^{y}}=1-y' \\ \Rightarrow y'=\dfrac{1-{{e}^{y}}}{1+x{{e}^{y}}} \\ \text{When }x=0, \\ \therefore y'=1-{{e}^{y}} $

 Find the implicit derivative with respect to $y$ of the function $4{{x}^{4}}-7{{y}^{3}}=\sqrt{6}x$.

 Apply the derivative of all the terms separately with respect to $y$ instead of $x$. Use formula derivative of product formula. Bring all the like terms to one term and get the required value. 

 $ \dfrac{d}{dy}\left( 4{{x}^{4}} \right)-\dfrac{d}{dy}\left( 7{{y}^{3}} \right)=\dfrac{d}{dy}\left( \sqrt{6}x \right) \\ \Rightarrow 16{{x}^{3}}\dfrac{dx}{dy}-21{{y}^{2}}=\sqrt{6}\dfrac{dx}{dy} \\ \Rightarrow \left( 16{{x}^{3}}-\sqrt{6} \right)\dfrac{dx}{dy}=21{{y}^{2}} \\ \Rightarrow \dfrac{dx}{dy}=-\dfrac{21{{y}^{2}}}{\left( \sqrt{6}-16{{x}^{3}} \right)} $

 Find $y''$, of $y=x{{y}^{2}}+xy$ when $x=0$ and $y=1$.

 Apply the derivative of all the terms separately with respect to $x$, here $\dfrac{dy}{dx}=y'$. Use formula derivative of product formula. Bring all the like terms to one term and get the value of $y'$.

$ \dfrac{d}{dx}\left( y \right)=\dfrac{d}{dx}\left( x{{y}^{2}} \right)+\dfrac{d}{dx}\left( xy \right) \\ \Rightarrow y'=x\dfrac{d}{dx}\left( {{y}^{2}} \right)+{{y}^{2}}+x\dfrac{d}{dx}\left( y \right)+y \\ \Rightarrow y'=2xyy'+{{y}^{2}}+xy'+y.......(i) \\ \therefore y'=\dfrac{{{y}^{2}}+y}{(1-2xy-x)} \\ \text{When }x=0,y=1, \\ y'=2........(ii) \\ \text{Differentiating equation (i), we get } \\ y^{\prime \prime}=2\left[xy\dfrac{d}{d x}(y^\prime)+x y^{\prime} \dfrac{d}{d x}(y)+yy^\prime \dfrac{d}{d x}\left(x\right)\right]+\dfrac{d}{d x}\left(y^{2}\right)+x \dfrac{d}{d x}\left(y^{\prime}\right)+y^\prime \dfrac{d}{dx}(x)+\dfrac{d}{dx}(y) \\ y^{\prime \prime}=2\left[xyy^{\prime\prime}+x(y^{\prime})^2+2yy^{\prime}+y^{\prime}\right]+xy^{\prime\prime} \\ \text{when }x=0, y=1 \text{ and }y^{\prime}=2\text{ we get,} \\ 2[2(2)+2]=12$