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Find \[\dfrac{du}{dv}\] at \[\theta =\dfrac{\pi }{4}\] when \[u=\log \left( \sec \theta +\tan \theta \right);v={{e}^{\left( \cos \theta -\sin \theta \right)}}\]

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Last updated date: 16th Jul 2024
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Answer
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Hint: In this type of question we have to use the concept of derivatives. Here, we have given two functions \[u\] and \[v\], where both are functions of \[\theta \]. So, first we differentiate both the functions with respect to \[\theta \] and then we find the value of \[\dfrac{du}{dv}\] by using, \[\dfrac{du}{dv}=\dfrac{\left( \dfrac{du}{d\theta } \right)}{\left( \dfrac{dv}{d\theta } \right)}\]. Also we have to substitute \[\theta =\dfrac{\pi }{4}\] to obtain the final result.

Complete step-by-step solution:
Now, we have to find \[\dfrac{du}{dv}\] at \[\theta =\dfrac{\pi }{4}\] when \[u=\log \left( \sec \theta +\tan \theta \right);v={{e}^{\left( \cos \theta -\sin \theta \right)}}\].
Let us consider,
\[\Rightarrow u=\log \left( \sec \theta +\tan \theta \right)\]
Now, we differentiate \[u\] with respect to \[\theta \],
\[\Rightarrow \dfrac{du}{d\theta }=\dfrac{d}{d\theta }\left( \log \left( \sec \theta +\tan \theta \right) \right)\]
As we know that, \[\dfrac{d}{dx}\log x=\dfrac{1}{x}\]
\[\Rightarrow \dfrac{du}{d\theta }=\dfrac{1}{\left( \sec \theta +\tan \theta \right)}\dfrac{d}{d\theta }\left( \sec \theta +\tan \theta \right)\]
Also we know that, \[\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x,\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x\]
\[\Rightarrow \dfrac{du}{d\theta }=\dfrac{1}{\left( \sec \theta +\tan \theta \right)}\left( \sec \theta \tan \theta +{{\sec }^{2}}\theta \right)\]
\[\begin{align}
  & \Rightarrow \dfrac{du}{d\theta }=\dfrac{\sec \theta }{\left( \sec \theta +\tan \theta \right)}\left( \tan \theta +\sec \theta \right) \\
 & \Rightarrow \dfrac{du}{d\theta }=\sec \theta \\
\end{align}\]
\[\Rightarrow \dfrac{du}{d\theta }=\dfrac{1}{\cos \theta }\]
Now, let us consider,
\[\Rightarrow v={{e}^{\left( \cos \theta -\sin \theta \right)}}\]
By differentiating with respect to \[\theta \], we get,
\[\Rightarrow \dfrac{dv}{d\theta }=\dfrac{d}{d\theta }\left( {{e}^{\left( \cos \theta -\sin \theta \right)}} \right)\]
As we know that, \[\dfrac{d}{dx}{{e}^{x}}={{e}^{x}},\dfrac{d}{dx}\cos x=-\sin x,\dfrac{d}{dx}\sin x=\cos x\]
\[\begin{align}
  & \Rightarrow \dfrac{dv}{d\theta }={{e}^{\left( \cos \theta -\sin \theta \right)}}\dfrac{d}{d\theta }\left( \cos \theta -\sin \theta \right) \\
 & \Rightarrow \dfrac{dv}{d\theta }={{e}^{\left( \cos \theta -\sin \theta \right)}}\left( -\sin \theta -\cos \theta \right) \\
 & \Rightarrow \dfrac{dv}{d\theta }=-{{e}^{\left( \cos \theta -\sin \theta \right)}}\left( \sin \theta +\cos \theta \right) \\
\end{align}\]
Now, we have to find the value of \[\dfrac{du}{dv}\], for that let us consider,
\[\Rightarrow \dfrac{du}{dv}=\dfrac{\left( \dfrac{du}{d\theta} \right)}{\left( \dfrac{dv}{d\theta } \right)}\]
By substituting the values of \[\dfrac{du}{d\theta }\] and \[\dfrac{dv}{d\theta }\] in above equation we can write,
\[\Rightarrow \dfrac{du}{dv}=\dfrac{\left( \dfrac{1}{\cos \theta } \right)}{\left( -{{e}^{\left( \cos \theta -\sin \theta \right)}}\left( \sin \theta +\cos \theta \right) \right)}\]
Now substitute \[\theta =\dfrac{\pi }{4}\], hence we get,
\[\Rightarrow {{\left( \dfrac{du}{dv} \right)}_{\left( \theta =\dfrac{\pi }{4} \right)}}=\dfrac{\left( \dfrac{1}{\cos \dfrac{\pi }{4}} \right)}{\left( -{{e}^{\left( \cos \dfrac{\pi }{4}-\sin \dfrac{\pi }{4} \right)}}\left( \sin \dfrac{\pi }{4}+\cos \dfrac{\pi }{4} \right) \right)}\]
As we have the values of \[\cos \dfrac{\pi }{4}=\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] we can write,
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{du}{dv} \right)}_{\left( \theta =\dfrac{\pi }{4} \right)}}=\dfrac{\left( \dfrac{1}{\dfrac{1}{\sqrt{2}}} \right)}{\left( -{{e}^{\left( \dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}} \right)}}\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \right) \right)} \\
 & \Rightarrow {{\left( \dfrac{du}{dv} \right)}_{\left( \theta =\dfrac{\pi }{4} \right)}}=\dfrac{\left( \sqrt{2} \right)}{\left( -{{e}^{0}}\left( \dfrac{2}{\sqrt{2}} \right) \right)} \\
\end{align}\]
We know that, \[{{e}^{0}}=1\]
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{du}{dv} \right)}_{\left( \theta =\dfrac{\pi }{4} \right)}}=\dfrac{\left( \sqrt{2} \right)}{\left( -\sqrt{2} \right)} \\
 & \Rightarrow {{\left( \dfrac{du}{dv} \right)}_{\left( \theta =\dfrac{\pi }{4} \right)}}=-1 \\
\end{align}\]
Hence, the value of \[\dfrac{du}{dv}\] at \[\theta =\dfrac{\pi }{4}\] when \[u=\log \left( \sec \theta +\tan \theta \right);v={{e}^{\left( \cos \theta -\sin \theta \right)}}\] is \[-1\].

Note: In this type of question students have to note that they have to first differentiate both the given functions first with respect to \[\theta \] and then can find the value of \[\dfrac{du}{dv}\]. Students have to remember the formulas of derivatives for the trigonometric functions as well as for exponential and logarithmic functions. Also students have to remember the values of trigonometric ratios for the angle \[\theta =\dfrac{\pi }{4}\]. Students have to take care during simplification after substituting the value of \[\theta =\dfrac{\pi }{4}\].