Questions & Answers

Question

Answers

Answer
Verified

Hint: For solving this question we will use the product rule of differentiation for differentiating any function of the form $y=f\left( x \right)\cdot g\left( x \right)$. After that, we will use the result of derivative of ${{e}^{x}}$ and $\tan x$ with respect to $x$ for writing the derivative of ${{e}^{x}}\tan x$ with respect to $x$.

Complete step-by-step answer:

We have a function $y={{e}^{x}}\tan x$ and we have to find the derivative of the given function with respect to the variable $x$ .

Now, before we proceed we should know the following three results:

1. If $y=f\left( x \right)\cdot g\left( x \right)$ then derivative of $y$ with respect to the variable $x$ will be equal to $\dfrac{dy}{dx}={f}'\left( x \right)\cdot g\left( x \right)+f\left( x \right)\cdot {g}'\left( x \right)$ . This is also known as the product rule of differentiation.

2. If $y={{e}^{x}}$ then derivative of $y$ with respect to the variable $x$ will be equal to $\dfrac{dy}{dx}=\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ .

3. If $y=\tan x$ then derivative of $y$ with respect to the variable $x$ will be equal to $\dfrac{dy}{dx}=\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$ .

Now, we will be using the above three results for solving this question. As it is given that $y={{e}^{x}}\tan x$ and we have to find the derivative of the given function with respect to the variable $x$ so, we can apply the product rule of differentiation which is mentioned in the first point. Then,

$\begin{align}

& y={{e}^{x}}\tan x \\

& \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{e}^{x}} \right)}{dx}\times \tan x+{{e}^{x}}\times \dfrac{d\left( \tan x \right)}{dx} \\

\end{align}$

Now, form the second and third results mentioned above we can substitute $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ and $\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$ in the above equation. Then,

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{d\left( {{e}^{x}} \right)}{dx}\times \tan x+{{e}^{x}}\times \dfrac{d\left( \tan x \right)}{dx} \\

& \Rightarrow \dfrac{dy}{dx}={{e}^{x}}\tan x+{{e}^{x}}{{\sec }^{2}}x \\

& \Rightarrow \dfrac{dy}{dx}={{e}^{x}}\left( \tan x+{{\sec }^{2}}x \right) \\

\end{align}$

Now, from the above result, we can write that if $y={{e}^{x}}\tan x$ then derivative of the given function with respect to the variable $x$ will be equal to $\dfrac{dy}{dx}={{e}^{x}}\left( \tan x+{{\sec }^{2}}x \right)$ .

Thus, differentiation of ${{e}^{x}}\tan x$ with respect to $x$ will be equal to ${{e}^{x}}\left( \tan x+{{\sec }^{2}}x \right)$.

Note: Here, the student should understand how we should apply the product rule to find the value of $\dfrac{dy}{dx}$ for any function of the form $y=f\left( x \right)\cdot g\left( x \right)$ . Although the problem is very easy, we should be careful while solving. Moreover, the student should avoid calculation mistakes while solving to get the correct answer.

Complete step-by-step answer:

We have a function $y={{e}^{x}}\tan x$ and we have to find the derivative of the given function with respect to the variable $x$ .

Now, before we proceed we should know the following three results:

1. If $y=f\left( x \right)\cdot g\left( x \right)$ then derivative of $y$ with respect to the variable $x$ will be equal to $\dfrac{dy}{dx}={f}'\left( x \right)\cdot g\left( x \right)+f\left( x \right)\cdot {g}'\left( x \right)$ . This is also known as the product rule of differentiation.

2. If $y={{e}^{x}}$ then derivative of $y$ with respect to the variable $x$ will be equal to $\dfrac{dy}{dx}=\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ .

3. If $y=\tan x$ then derivative of $y$ with respect to the variable $x$ will be equal to $\dfrac{dy}{dx}=\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$ .

Now, we will be using the above three results for solving this question. As it is given that $y={{e}^{x}}\tan x$ and we have to find the derivative of the given function with respect to the variable $x$ so, we can apply the product rule of differentiation which is mentioned in the first point. Then,

$\begin{align}

& y={{e}^{x}}\tan x \\

& \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{e}^{x}} \right)}{dx}\times \tan x+{{e}^{x}}\times \dfrac{d\left( \tan x \right)}{dx} \\

\end{align}$

Now, form the second and third results mentioned above we can substitute $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ and $\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$ in the above equation. Then,

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{d\left( {{e}^{x}} \right)}{dx}\times \tan x+{{e}^{x}}\times \dfrac{d\left( \tan x \right)}{dx} \\

& \Rightarrow \dfrac{dy}{dx}={{e}^{x}}\tan x+{{e}^{x}}{{\sec }^{2}}x \\

& \Rightarrow \dfrac{dy}{dx}={{e}^{x}}\left( \tan x+{{\sec }^{2}}x \right) \\

\end{align}$

Now, from the above result, we can write that if $y={{e}^{x}}\tan x$ then derivative of the given function with respect to the variable $x$ will be equal to $\dfrac{dy}{dx}={{e}^{x}}\left( \tan x+{{\sec }^{2}}x \right)$ .

Thus, differentiation of ${{e}^{x}}\tan x$ with respect to $x$ will be equal to ${{e}^{x}}\left( \tan x+{{\sec }^{2}}x \right)$.

Note: Here, the student should understand how we should apply the product rule to find the value of $\dfrac{dy}{dx}$ for any function of the form $y=f\left( x \right)\cdot g\left( x \right)$ . Although the problem is very easy, we should be careful while solving. Moreover, the student should avoid calculation mistakes while solving to get the correct answer.

×

Sorry!, This page is not available for now to bookmark.