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

Trigonometric derivatives Quiz 2

mathematics

 Find  $\varphi '\left( x \right)\left( 1-{{x}^{2}} \right)-\varphi \left( x \right)x$ if  $\varphi \left( x \right)=\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$.

 First cross multiply and square both sides of the given expression and then differentiate both sides w.r.t $x$.

  $\varphi \left( x \right)=\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} \\  \Rightarrow \left( 1-{{x}^{2}} \right){{\left[ \varphi \left( x \right) \right]}^{2}}={{\left( {{\sin }^{-1}}x \right)}^{2}} \\  \text{Differentiating both sides w}\text{.r}\text{.t }x \\  \Rightarrow \left( 1-{{x}^{2}} \right)2\varphi \left( x \right)\varphi'\left( x \right)+{{\left[ \varphi \left( x \right) \right]}^{2}}\left( -2x \right)=\dfrac{2{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} \\  \Rightarrow \left( 1-{{x}^{2}} \right)2\varphi \left( x \right)\varphi '\left( x \right)+{{\left[ \varphi \left( x \right) \right]}^{2}}\left( -2x \right)=2\varphi \left( x \right) \\  \Rightarrow \left( 1-{{x}^{2}} \right)\varphi '\left( x \right)-\left[ \varphi \left( x \right) \right]\left( x \right)=1 \\ $

 Find the slope of normal to the curve  $y={{\sin }^{-1}}\left( \dfrac{\log {{x}^{2}}}{1+{{\left( \log x \right)}^{2}}} \right)$.

 Simplify the given expression by assuming $\log x=\tan \phi $, it would reduce the given expression to $y=2\phi $. Then, use the relation between the slopes of tangent and normal i.e. ${{\text{m}}_{1}}{{\text{m}}_{2}}=-1$.

  $ y={{\sin }^{-1}}\left( \dfrac{\log {{x}^{2}}}{1+{{\left( \log x \right)}^{2}}} \right) \\  \Rightarrow y={{\sin }^{-1}}\left( \dfrac{2\log x}{1+{{\left( \log x \right)}^{2}}} \right) \\  \text{Let }\log x=\tan \phi  \\  \Rightarrow y={{\sin }^{-1}}\left( \dfrac{2\tan \phi }{1+{{\tan }^{2}}\phi } \right)=2\phi  \\  \Rightarrow y=2{{\tan }^{-1}}\left( \log x \right) \\  \Rightarrow \dfrac{dy}{dx}=\dfrac{2}{1+{{\left( \log x \right)}^{2}}}\times \dfrac{1}{x}=\dfrac{2}{x\left[ 1+{{\left( \log x \right)}^{2}} \right]} \\  \text{Now }\dfrac{dy}{dx}\text{ represents the slope of tangent}\text{.} \\  \text{Slope of normal}=-\dfrac{x\left[ 1+{{\left( \log x \right)}^{2}} \right]}{2}\left( \because \text{For perpendicular lines }{{\text{m}}_{1}}{{\text{m}}_{2}}=-1 \right) \\ $

 If  $\varphi \left( h \right)=\sinh -\cosh $ find the interval in which the function is decreasing in  $0\le h\le 2\pi $. 

 First differentiate the given function and then convert the obtained function into cosine function by multiplying and dividing by  $\sqrt{2}$. Now, take $\phi '\left( h \right) < 0$ and solve for the required answer.

$ \varphi \left( h \right)=\sinh -\cosh  \\  \text{Differentiating both sides w}\text{.r}\text{.t ''h''} \\  \Rightarrow \varphi '\left( h \right)=\cosh +\sinh  \\  \Rightarrow \varphi '\left( h \right)=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\cosh +\dfrac{1}{\sqrt{2}}\sinh  \right) \\  \Rightarrow \varphi '\left( h \right)=\sqrt{2}\left( \cosh \cos \dfrac{\pi }{4}+\sinh \sin \dfrac{\pi }{4} \right) \\  \Rightarrow \varphi '\left( h \right)=\sqrt{2}\left( \cos \left( h-\dfrac{\pi }{4} \right) \right)\text{  }\left[ \cos (A-B)=\cos A\cos B+\sin A\sin B \right] \\  \text{For decreasing }\varphi '\left( h \right) < 0 \\  \Rightarrow \dfrac{\pi }{2} < \left( h-\dfrac{\pi }{4} \right) < \dfrac{3\pi }{2}\text{             }\left[ \because \text{ cos is negative in 2nd and 3rd quadrants} \right] \\  \Rightarrow \dfrac{3\pi }{4} < h < \dfrac{7\pi }{4} \\ $

 Find  $\dfrac{dy}{dx}$ if ${{e}^{xy}}+\log \left( xy \right)+\cos \left( xy \right)+15=0$.

 Differentiate using the chain rule i.e.  $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$ and also use the relations $\dfrac{d}{dx}\log x=\dfrac{1}{x}$  and  $\dfrac{d}{dx}\cos x=-\sin x$.

 $ {{e}^{xy}}+\log \left( xy \right)+\cos \left( xy \right)+15=0 \\  \Rightarrow {{e}^{xy}}\dfrac{d}{dx}\left( xy \right)+\dfrac{1}{xy}\dfrac{d}{dx}\left( xy \right)-\sin \left( xy \right)\dfrac{d}{dx}\left( xy \right)=0 \\  \Rightarrow \dfrac{d}{dx}\left( xy \right)\left\{ {{e}^{xy}}+\dfrac{1}{xy}-\sin \left( xy \right) \right\}=0 \\  \Rightarrow \dfrac{d}{dx}\left( xy \right)=0\left( \because \text{ }{{e}^{xy}}+\dfrac{1}{xy}-\sin \left( xy \right)\ne 0 \right) \\  \Rightarrow x\dfrac{dy}{dx}+y.1=0 \\  \Rightarrow \dfrac{dy}{dx}=-\dfrac{y}{x} \\ $

 Find $\dfrac{dy}{dx}$  at  $x=1$, if  $y=\phi \left( \dfrac{2x+3}{3-2x} \right)$ and  $\phi '\left( x \right)=\sin \left( \log x \right)$

 Differentiate both sides of $y=\phi \left( \dfrac{2x+3}{3-2x} \right)$ and then apply chain rule i.e., $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$. Now, use the relations $\dfrac{d}{dx}\log x=\dfrac{1}{x}$ and $\dfrac{d}{dx}\sin x=\cos x$.

 $ y=\phi \left( \dfrac{2x+3}{3-2x} \right) \\  \Rightarrow \dfrac{dy}{dx}=\phi '\left( \dfrac{2x+3}{3-2x} \right)\dfrac{d}{dx}\left( \dfrac{2x+3}{3-2x} \right) \\  \Rightarrow \dfrac{dy}{dx}=\sin \left[ \log \left( \dfrac{2x+3}{3-2x} \right) \right]\left( \dfrac{\left( 3-2x \right)\left( 2 \right)-\left( 2x+3 \right)\left( -2 \right)}{{{\left( 3-2x \right)}^{2}}} \right) \\  \text{          }=\dfrac{12}{{{\left( 3-2x \right)}^{2}}}\sin \left[ \log \left( \dfrac{2x+3}{3-2x} \right) \right] \\  \Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{x=1}}=\dfrac{12}{{{\left( 3-2 \right)}^{2}}}\sin \left[ \log \left( 5 \right) \right]=12\sin \log 5 \\ $