Solveeit Logo

Question

Question: Differentiate \({{e}^{x}}\tan x\) with respect to \(x\)....

Differentiate extanx{{e}^{x}}\tan x with respect to xx.

Explanation

Solution

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

Complete step-by-step answer:
We have a function y=extanxy={{e}^{x}}\tan x and we have to find the derivative of the given function with respect to the variable xx .
Now, before we proceed we should know the following three results:
1. If y=f(x)g(x)y=f\left( x \right)\cdot g\left( x \right) then derivative of yy with respect to the variable xx will be equal to dydx=f(x)g(x)+f(x)g(x)\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=exy={{e}^{x}} then derivative of yy with respect to the variable xx will be equal to dydx=d(ex)dx=ex\dfrac{dy}{dx}=\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}} .
3. If y=tanxy=\tan x then derivative of yy with respect to the variable xx will be equal to dydx=d(tanx)dx=sec2x\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=extanxy={{e}^{x}}\tan x and we have to find the derivative of the given function with respect to the variable xx so, we can apply the product rule of differentiation which is mentioned in the first point. Then,
y=extanx dydx=d(ex)dx×tanx+ex×d(tanx)dx \begin{aligned} & 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{aligned}
Now, form the second and third results mentioned above we can substitute d(ex)dx=ex\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}} and d(tanx)dx=sec2x\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x in the above equation. Then,
dydx=d(ex)dx×tanx+ex×d(tanx)dx dydx=extanx+exsec2x dydx=ex(tanx+sec2x) \begin{aligned} & \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{aligned}
Now, from the above result, we can write that if y=extanxy={{e}^{x}}\tan x then derivative of the given function with respect to the variable xx will be equal to dydx=ex(tanx+sec2x)\dfrac{dy}{dx}={{e}^{x}}\left( \tan x+{{\sec }^{2}}x \right) .
Thus, differentiation of extanx{{e}^{x}}\tan x with respect to xx will be equal to ex(tanx+sec2x){{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 dydx\dfrac{dy}{dx} for any function of the form y=f(x)g(x)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.