Question

Differentiate the following with respect to \[x\].
1.\[\dfrac{{{e}^{x}}}{\sin x}\]
2. \[\sin \left( {{\tan }^{-1}}{{e}^{-x}} \right)\]
3. \[\sqrt{{{e}^{\sqrt{x}}}},x>0\]

Answer 1

Hint: We will be considering each of the three functions and then we must be differentiating with respect to \[x\], by applying the apt method for differentiating. The obtained answer of each of the function would be our required answer

Complete step-by-step solution:
Now let us learn about differentiation. Differentiation is a method which is used to find the instantaneous rate of change of function. It is used in finding the derivative of a function. Differentiation can be done even by applying the limits to the functions. There are four basic rules of differentiation. They are: sum and the difference rule, product rule, quotient rule and the chain rule.
Now let us start differentiating the given functions.
Firstly, let us consider, \[\dfrac{{{e}^{x}}}{\sin x}\]
Since we can see that it is in the form of \[\dfrac{u}{v}\], we will be applying the \[\dfrac{u}{v}\] method.
So we have, \[u={{e}^{x}}\] and \[v=\sin x\].
Let us consider it as \[y=\dfrac{{{e}^{x}}}{\sin x}\]
\[\Rightarrow y=\dfrac{u}{v}\]
Now upon differentiating on both sides with respect to \[x\], we get
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d\left( \dfrac{u}{v} \right)}{dx} \\
 & \dfrac{dy}{dx}=\dfrac{{{u}^{'}}v-{{v}^{'}}u}{{{v}^{2}}} \\
\end{align}\]
Now let us substitute the respective values. We get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{d\left( {{e}^{x}} \right)}{dx}\centerdot \sin x-\dfrac{d\left( \sin x \right)}{dx}\centerdot {{e}^{x}}}{{{\sin }^{2}}x}\]
Now let us differentiate it. We obtain,
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{d\left( {{e}^{x}} \right)}{dx}\centerdot \sin x-\dfrac{d\left( \sin x \right)}{dx}\centerdot {{e}^{x}}}{{{\sin }^{2}}x} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{e}^{x}}\centerdot \sin x-\cos x\centerdot {{e}^{x}}}{{{\sin }^{2}}x} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{e}^{x}}\left( \sin x-\cos x \right)}{{{\sin }^{2}}x} \\
\end{align}\]
\[\therefore \] The derivative of \[\dfrac{{{e}^{x}}}{\sin x}\] is \[\dfrac{{{e}^{x}}\left( \sin x-\cos x \right)}{{{\sin }^{2}}x}\].
Now let us consider the next function \[\sin \left( {{\tan }^{-1}}{{e}^{-x}} \right)\].
We have \[y=\sin \left( {{\tan }^{-1}}{{e}^{-x}} \right)\]
Upon differentiating on both sides with respect to \[x\] we get,
\[{{y}^{'}}={{\left( \sin \left( {{\tan }^{-1}}{{e}^{-x}} \right) \right)}^{'}}\]
\[\begin{align}
  & {{y}^{'}}={{\left( \sin \left( {{\tan }^{-1}}{{e}^{-x}} \right) \right)}^{'}} \\
 & \Rightarrow {{y}^{'}}= \cos \left( {{\tan }^{-1}}{{e}^{-x}} \right)\times {{\left( {{\tan }^{-1}}{{e}^{-x}} \right)}^{'}}\left[ \because {{\left( \text{sinx} \right)}^{\text{ }\!\!'\!\!\text{ }}}\text{=cosx} \right] \\
 & \Rightarrow {{y}^{'}}= \cos \left( {{\tan }^{-1}}{{e}^{-x}} \right)\times \dfrac{1}{1+{{\left( {{e}^{-x}} \right)}^{2}}}\times {{\left( {{e}^{-x}} \right)}^{'}}\left[ \because {{\left( {{\tan }^{-1}}x \right)}^{'}}=\dfrac{1}{1+{{x}^{2}}} \right] \\
 & \Rightarrow {{y}^{'}}= \cos \left( {{\tan }^{-1}}{{e}^{-x}} \right)\times \dfrac{1}{1+{{\left( {{e}^{-x}} \right)}^{2}}}\times -{{e}^{-x}} \\
 & \Rightarrow {{y}^{'}}= - \dfrac{{{e}^{-x}}\cos \left( {{\tan }^{-1}}{{e}^{-x}} \right)}{1+{{\left( {{e}^{-x}} \right)}^{2}}} \\
 & \Rightarrow {{y}^{'}}= -\dfrac{{{e}^{-x}}\cos \left( {{\tan }^{-1}}{{e}^{-x}} \right)}{1+{{e}^{-2x}}} \\
\end{align}\]
\[\therefore \] The derivative of \[\sin \left( {{\tan }^{-1}}{{e}^{-x}} \right)\] is \[- \dfrac{{{e}^{-x}}\cos \left( {{\tan }^{-1}}{{e}^{-x}} \right)}{1+{{e}^{-2x}}}\].
Now let us consider our next function \[\sqrt{{{e}^{\sqrt{x}}}},x>0\].
We have \[y=\sqrt{{{e}^{\sqrt{x}}}}\]
Upon differentiating on both sides with respect to \[x\] by applying the chain rule, we get,
\[\begin{align}
  & {{y}^{'}}={{\left( \sqrt{{{e}^{\sqrt{x}}}} \right)}^{'}} \\
 & \Rightarrow {{y}^{'}}=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}\times {{\left( {{e}^{\sqrt{x}}} \right)}^{'}} \\
 & \Rightarrow {{y}^{'}}=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}\times {{e}^{\sqrt{x}}}\times {{\left( \sqrt{x} \right)}^{'}} \\
 & \Rightarrow {{y}^{'}}=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}\times {{e}^{\sqrt{x}}}\times \dfrac{1}{2\sqrt{x}} \\
 & \Rightarrow {{y}^{'}}=\dfrac{{{e}^{\sqrt{x}}}}{4\sqrt{x\centerdot \sqrt{{{e}^{\sqrt{x}}}}}} \\
\end{align}\]
\[\therefore \] The derivative of \[\sqrt{{{e}^{\sqrt{x}}}}\] is \[\dfrac{{{e}^{\sqrt{x}}}}{4\sqrt{x\centerdot \sqrt{{{e}^{\sqrt{x}}}}}}\].

Note: We must always have a note regarding the variable that is to be considered for differentiation. We can apply the concept of differentiation in finding out the displacement with respect to time. The opposite of finding the derivative is called an antiderivative.