Question

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

Answer 1

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 $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}$. Let’s assume $u = {e^{\tan x}}$ and $v = \tan x$ to solve easily. Then use basic differentiation formulas such as $\dfrac{d}{{dx}}(\tan x) = {\sec ^2}x$, $\dfrac{d}{{dx}}({e^x}) = {e^x}$. After getting the differential form, put x=0 in the equation along with ${f'}(1) = 5$. 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 $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}$.
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,
$\Rightarrow$ $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(f(u))$
Now using the chain rule of differential calculus,
$\Rightarrow$ $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(f(u)) = \dfrac{d}{{du}}(f(u)) \cdot \dfrac{{du}}{{dx}}$
As we know from basic function of derivative $\dfrac{d}{{du}}(f(u)) = {f'}(u)$
So, $\dfrac{{dy}}{{dx}} = {f'}(u) \cdot \dfrac{{du}}{{dx}}$
Now, $u = {e^{\tan x}} = {e^v}$
Differentiating the above equation with respect to x,
$\Rightarrow$ $\dfrac{d}{{dx}}(u) = \dfrac{d}{{dx}}({e^v})$
Again using chain rule,
$\Rightarrow$ $\dfrac{d}{{dx}}({e^v}) = \dfrac{d}{{dv}}({e^v}) \cdot \dfrac{{dv}}{{dx}}$
Put $v = \tan x$ in above equation,
$\Rightarrow$ $\dfrac{d}{{dx}}({e^v}) = \dfrac{d}{{dv}}({e^v}) \cdot \dfrac{d}{{dx}}(\tan x)$
As we know that $\dfrac{d}{{dx}}({e^x}) = {e^x}$ and $\dfrac{d}{{dx}}(\tan x) = {\sec ^2}x$
So simplifying,
$\Rightarrow$ $\dfrac{d}{{dx}}({e^v}) = {e^v} \cdot {\sec ^2}x$.
Put, $v = \tan x$ in above equation,
So, $\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}({e^v}) = {e^{\tan x}} \cdot {\sec ^2}x$
Now in equation $\dfrac{{dy}}{{dx}} = {f'}(u) \cdot \dfrac{{du}}{{dx}}$, put value of $\dfrac{{du}}{{dx}}$.
So, $\dfrac{{dy}}{{dx}} = {f'}(u) \cdot {e^v} = {f'}(u) \cdot {e^{\tan x}} \cdot {\sec ^2}x$.
Putting, $u = {e^{\tan x}}$ in above equation,
$\Rightarrow$ $\dfrac{{dy}}{{dx}} = {f'}({e^{\tan x}}) \cdot {e^{\tan x}} \cdot {\sec ^2}x$
This equation is the derivative form of function $f({e^{\tan x}})$ with respect to x.
Now, for given condition at x=0, ${f'}(1) = 5$
Putting x=0 in final derivative form,
$\Rightarrow$ ${\dfrac{{dy}}{{dx}}_{x = 0}} = {f'}({e^{\tan 0}}) \cdot {e^{\tan 0}} \cdot {\sec ^2}0$
As we know that $\tan 0 = 0$ and $\sec 0 = 1$,
So,
$\Rightarrow$ ${\dfrac{{dy}}{{dx}}_{x = 0}} = {f'}({e^0}) \cdot {e^0} \cdot {(1)^2}$
And ${e^0} = 1$,
So, ${\dfrac{{dy}}{{dx}}_{x = 0}} = {f'}(1) \cdot 1 \cdot 1$
$\Rightarrow$ ${\dfrac{{dy}}{{dx}}_{x = 0}} = {f'}(1)$
Now given that ${f'}(1) = 5$ So, ${\dfrac{{dy}}{{dx}}_{x = 0}} = {f'}(1) = 5$

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, $\dfrac{d}{{dx}}(f(x) \cdot g(x)) = (\dfrac{d}{{dx}}f(x)) \cdot g(x) + f(x) \cdot (\dfrac{d}{{dx}}g(x))$. If the question is like $\tan x \cdot f({e^x})$ then we have to use the product rule of derivation.