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This is one of our many quizzes on Logarithmic derivatives. Click on this to access more such competitive quizzes on Differential Calculus.

Logarithmic derivatives Quiz 2

mathematics

 Differentiate $y=\sin \left( \log x+{{e}^{x}} \right).$

 First, assume $u=\left( \log x+{{e}^{x}} \right)$ and substitute secondly use $\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$ to get the derivative.

$y=\sin \left( \log x+{{e}^{x}} \right)$ Now $u=\left( \log x+{{e}^{x}} \right)$ $\Rightarrow \dfrac{du}{dx}=\dfrac{1}{x}+{{e}^{x}}$ $\Rightarrow \dfrac{dy}{du}=\cos u$ Now $\dfrac{dy}{dx}=\cos \left( \log x+{{e}^{x}} \right)\left( \dfrac{1}{x}+{{e}^{x}} \right)$.

 Differentiate the function $y=\sin \left[ \sin \left( \log 5x \right) \right]$.

 Derivative of $\log x$ is $\dfrac{1}{x}$ and derivative of $\sin x$ is $\cos x$. Also use chain rule for derivatives of the function. 

$y=\sin \left[ \sin \left( \log 5x \right) \right]$ $\dfrac{dy}{dx}=\cos \left[ \sin \left( \log 5x \right) \right]\dfrac{d}{dx}\sin \left( \log 5x \right)$ $\dfrac{dy}{dx}=\cos \left[ \sin \left( \log 5x \right) \right]\cos \left( \log 5x \right)\dfrac{d}{dx}\log 5x$ $\dfrac{dy}{dx}=\cos \left[ \sin \left( \log 5x \right) \right]\cos \left( \log 5x \right)\dfrac{1}{5x}.5$

 Differentiate with respect to $y$ $x={{\log }_{y}}19$.

 Use the identity ${{\log }_{a}}b=\dfrac{1}{{{\log }_{b}}a}$ in $x={{\left( {{\log }_{19}}y \right)}^{-1}}$ and $\dfrac{d}{dx}{{\log }_{a}}x=\dfrac{1}{x}{{\log }_{e}}e$ while differentiating the function.

&nbsp;$x={{\log }_{y}}19$ $\Rightarrow x={{\left( {{\log }_{19}}y \right)}^{-1}}$ $\Rightarrow \dfrac{dx}{dy}=-\dfrac{1}{{{\left( {{\log }_{19}}y \right)}^{2}}}\dfrac{d}{dy}{{\log }_{19}}y$ $\Rightarrow \dfrac{dx}{dy}=-\dfrac{1}{{{\left( {{\log }_{19}}y \right)}^{2}}}.\dfrac{1}{y{{\log }_{e}}19}$

 Differentiate $y={{e}^{\left( 2x+1 \right)}}\log \left[ \cos \left( 2x+1 \right) \right]$.

 Use multiplication rule for differentiation i.e. $\dfrac{d}{dx}u.v=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$.

$y={{e}^{2x+1}}\log \left[ \cos \left( 2x+1 \right) \right]$ $\Rightarrow \dfrac{dy}{dx}=2{{e}^{2x+1}}\log \left[ \cos \left( 2x+1 \right)+{{e}^{2x+1}}\dfrac{d}{dx}\log \left[ \cos \left( 2x+1 \right) \right] \right]$ $\Rightarrow 2{{e}^{2x+1}}\log \left[ \cos \left( 2x+1 \right) \right]+{{e}^{2x+1}}\dfrac{1}{\cos \left( 2x+1 \right)}\left[ -\sin \left( 2x+1 \right) \right].2$ $\Rightarrow 2{{e}^{2x+1}}\left[ \log \left[ \cos \left( 2x+1 \right) \right]-\tan \left( 2x+1 \right) \right]$

 Differentiate the function $\cos \left[ \cos \left( \log 7x \right) \right]$.

 Use chain rule in differentiation and $\dfrac{d}{dx}\log x=\dfrac{1}{x}$ for solving the differentiation of the function. 

$y=\cos \left[ \cos \left( \log 7x \right) \right]$ $\Rightarrow \dfrac{dy}{dx}=-\sin \left[ \cos \left( \log 7x \right) \right]\dfrac{d}{dx}\cos \left( \log 7x \right)$ $\Rightarrow \sin \left[ \cos \left( \log 7x \right) \right]\sin \left( \log 7x \right)\dfrac{d}{dx}\log 7x\,$ $\Rightarrow \sin \left[ \cos \left( \log 7x \right) \right]\sin \left( \log 7x \right)\dfrac{1}{x}$