Question

Find the derivative of $f(e^{\tan x})$ with respect to x at x=0, It is given that $f^{\prime }(1)=5$.

Answer 1

Hint: This type of question can be solved by using basic differentiation formulas. Here the given function $f(e^{\tan x})$ is a composite function. To obtain differentiation of composite function use the chain rule of differential calculus. Chain rule of differential calculus is $${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{dy}{du}}\cdot {\displaystyle \frac{du}{dx}}$$. Let’s assume $u=e^{\tan x}$ and $v=\tan x$ to solve easily. Then use basic differentiation formulas such as ${\displaystyle \frac{d}{dx}}(\tan x)={\sec }^{2}x$, ${\displaystyle \frac{d}{dx}}(e^x)=e^x$. After getting the differential form, put x=0 in the equation along with $f^{\prime }(1)=5$. We will obtain a derivative of $f(e^{\tan x})$.

Complete step-by-step answer:
Here the given function is $f(e^{\tan x})$.
Let’s say $y=f(e^{\tan x})$.
Here the given function y is a composite function of calculus.
Derivative of the composite function can be obtained by using the chain rule of differential calculus.
Chain rule of differential calculus is given by $${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{dy}{du}}\cdot {\displaystyle \frac{du}{dx}}$$.
Now for the given function $y=f(e^{\tan x})$, let’s assume $u=e^{\tan x}$ and $v=\tan x$.
So, $u=e^{\tan x}=e^v$
And,$y=f(e^{\tan x})=f(u)$.
Taking derivative of function y with respect to x on both side of the equation,
$\Rightarrow$${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d}{dx}}(f(u))$
Now using the chain rule of differential calculus,
$\Rightarrow$${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d}{dx}}(f(u))={\displaystyle \frac{d}{du}}(f(u))\cdot {\displaystyle \frac{du}{dx}}$
As we know from basic function of derivative ${\displaystyle \frac{d}{du}}(f(u))=f^{\prime }(u)$
So, ${\displaystyle \frac{dy}{dx}}=f^{\prime }(u)\cdot {\displaystyle \frac{du}{dx}}$
Now, $u=e^{\tan x}=e^v$
Differentiating the above equation with respect to x,
$\Rightarrow$${\displaystyle \frac{d}{dx}}(u)={\displaystyle \frac{d}{dx}}(e^v)$
Again using chain rule,
$\Rightarrow$${\displaystyle \frac{d}{dx}}(e^v)={\displaystyle \frac{d}{dv}}(e^v)\cdot {\displaystyle \frac{dv}{dx}}$
Put $v=\tan x$ in above equation,
$\Rightarrow$${\displaystyle \frac{d}{dx}}(e^v)={\displaystyle \frac{d}{dv}}(e^v)\cdot {\displaystyle \frac{d}{dx}}(\tan x)$
As we know that ${\displaystyle \frac{d}{dx}}(e^x)=e^x$ and ${\displaystyle \frac{d}{dx}}(\tan x)={\sec }^{2}x$
So simplifying,
$\Rightarrow$${\displaystyle \frac{d}{dx}}(e^v)=e^v\cdot {\sec }^{2}x$.
Put, $v=\tan x$ in above equation,
So, ${\displaystyle \frac{du}{dx}}={\displaystyle \frac{d}{dx}}(e^v)=e^{\tan x}\cdot {\sec }^{2}x$
Now in equation ${\displaystyle \frac{dy}{dx}}=f^{\prime }(u)\cdot {\displaystyle \frac{du}{dx}}$, put value of ${\displaystyle \frac{du}{dx}}$.
So, ${\displaystyle \frac{dy}{dx}}=f^{\prime }(u)\cdot e^v=f^{\prime }(u)\cdot e^{\tan x}\cdot {\sec }^{2}x$.
Putting, $u=e^{\tan x}$ in above equation,
$\Rightarrow$${\displaystyle \frac{dy}{dx}}=f^{\prime }(e^{\tan x})\cdot e^{\tan x}\cdot {\sec }^{2}x$
This equation is the derivative form of function $f(e^{\tan x})$ with respect to x.
Now, for given condition at x=0, $f^{\prime }(1)=5$
Putting x=0 in final derivative form,
$\Rightarrow$${{\displaystyle \frac{dy}{dx}}}_{x=0}=f^{\prime }(e^{\tan 0})\cdot e^{\tan 0}\cdot {\sec }^{2}0$
As we know that $\tan 0=0$ and $\sec 0=1$,
So,
$\Rightarrow$${{\displaystyle \frac{dy}{dx}}}_{x=0}=f^{\prime }(e^0)\cdot e^0\cdot (1{)}^2$
And $e^0=1$,
So, ${{\displaystyle \frac{dy}{dx}}}_{x=0}=f^{\prime }(1)\cdot 1\cdot 1$
$\Rightarrow$${{\displaystyle \frac{dy}{dx}}}_{x=0}=f^{\prime }(1)$
Now given that $f^{\prime }(1)=5$ So, ${{\displaystyle \frac{dy}{dx}}}_{x=0}=f^{\prime }(1)=5$

So, the derivative of the function $f(e^{\tan x})$ with respect to x at x=0, is 5.

Note: Derivative of the composite function $f(g(x))$ (function of function) can be solved by chain rule of derivation. But if a given function is multiplication of two functions like $f(x)\cdot g(x)$, then we have to use the product rule of derivation. Product rule of derivation is given by, ${\displaystyle \frac{d}{dx}}(f(x)\cdot g(x))=({\displaystyle \frac{d}{dx}}f(x))\cdot g(x)+f(x)\cdot ({\displaystyle \frac{d}{dx}}g(x))$. If the question is like $$\tan x\cdot f(e^x)$$ then we have to use the product rule of derivation.