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Question

Question: How do you find the derivative of \(y = {e^{5x}}\) ?...

How do you find the derivative of y=e5xy = {e^{5x}} ?

Explanation

Solution

Here the basic concept which we are going to use is using the chain rule. We will find the derivative with respect to x using the chain rule. We have to apply the chain rule here because there is some numerical value other than 1 in place of the coefficient of x.

Complete Step by Step Solution:
The given equation is y=e5xy = {e^{5x}}
Differentiating both sides with respect to x, we get
dydx=d(e5x)dx\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{e}^{5x}} \right)}{dx}
As we know the first derivative of ex{e^x} is ex{e^x}, but here in the equation, there is 5, present in place of the coefficient of x, so we have to apply the chain rule to find its derivative.
The chain rule states that the derivative of f(g(x))f\left( {g\left( x \right)} \right) is f(g(x))g(x)f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right) , helps us differentiate composite functions.
So, here first we will differentiate e5x{e^{5x}} with respect to x, and then we will differentiate   5x\;5x with respect to x.
dydx=e5xd(5x)dx\Rightarrow \dfrac{{dy}}{{dx}} = {e^{5x}} \cdot \dfrac{{d\left( {5x} \right)}}{{dx}}
dydx=e5x5\Rightarrow \dfrac{{dy}}{{dx}} = {e^{5x}} \cdot 5
Rewriting above equation

dydx=5e5x \Rightarrow \dfrac{{dy}}{{dx}} = 5{e^{5x}}

Additional Information:
The chain rule is very important as we have to use it in many of the problems related to finding derivatives of composite functions.

Note:
There is an alternative method to solve this by taking log on both sides
Given equation y=e5xy = {e^{5x}}
Take log both sides
logey=logee5x\Rightarrow {\log _e}y = {\log _e}{e^{5x}}
logey=5xlogee\Rightarrow {\log _e}y = 5x \cdot {\log _e}e
As we know that logee=1{\log _e}e = 1
Hence, logey=5x{\log _e}y = 5x
We know that d(logex)dx=1x\dfrac{{d\left( {{{\log }_e}x} \right)}}{{dx}} = \dfrac{1}{x}
So, now differentiating both sides with respect to x,
1y(dydx)=5\Rightarrow \dfrac{1}{y}\left( {\dfrac{{dy}}{{dx}}} \right) = 5
dydx=5y\Rightarrow \dfrac{{dy}}{{dx}} = 5y
As given in the question, y=e5xy = {e^{5x}} , therefore
dydx=5e5x\Rightarrow \dfrac{{dy}}{{dx}} = 5{e^{5x}}.