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Question

Question: How do you find the derivative of \(y={{e}^{2x}}\)?...

How do you find the derivative of y=e2xy={{e}^{2x}}?

Explanation

Solution

To get the derivative of y=e2xy={{e}^{2x}} with respect to xx. Firstly, suppose t=2xt=2x and get the derivative of tt with respect to xx . Now we can write y=e2xy={{e}^{2x}} as y=ety={{e}^{t}} and after that try to get the derivative with respect to tt. After combining both the derivative as dydx=dydt×dtdx\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx} we can get the derivative of y=e2xy={{e}^{2x}}with respect to xx .

Complete step by step solution:
The question has the given equation as y=e2xy={{e}^{2x}}
Since, we cannot derive the given equation directly. So, we would assume
t=2xt=2x(i)\left( i \right)
After that we have to derive equation (i)\left( i \right) with respect to xx as
dtdx=d(2x)dx\Rightarrow \dfrac{dt}{dx}=\dfrac{d\left( 2x \right)}{dx}
Since, numbers are constant in any derivation, so we cannot derive22 . Forxx, the derivation will be 11 .
So,
dtdx=2\Rightarrow \dfrac{dt}{dx}=2(ii)\left( ii \right)
With the use of equation (i)\left( i \right) , we can write the given equation in the question y=e2xy={{e}^{2x}} as:
y=et\Rightarrow y={{e}^{t}}
Since, the derivation of et{{e}^{t}} with respect to tt is itself et{{e}^{t}} . So, after derivation of the above equation with respect to tt will be as
dydt=et\Rightarrow \dfrac{dy}{dt}={{e}^{t}}
Now, after using equation (i)\left( i \right) , we can write the above derivation in term of xx as
dydt=\Rightarrow \dfrac{dy}{dt}= e2x{{e}^{2x}}(iii)\left( iii \right)
Now, for getting the derivative of y=e2xy={{e}^{2x}} with respect to xx , we can use the method
dydx=dydt×dtdx\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}(iv)\left( iv \right)
After applying the equation (ii)\left( ii \right) and (iii)\left( iii \right) in equation (iv)\left( iv \right) , we get
dydx=e2x×2\Rightarrow \dfrac{dy}{dx}={{e}^{2x}}\times 2
We can write the above equation as
dydx=2e2x\Rightarrow \dfrac{dy}{dx}=2{{e}^{2x}}
Hence, the derivative of the equation y=e2xy={{e}^{2x}} is 2e2x2{{e}^{2x}} .

Note:
Here we can check whether the derivative of the given equation is correct or not in the following way-
From the solution, we have:
dydx=2e2x\dfrac{dy}{dx}=2{{e}^{2x}}
We can write it as:
dy=(2e2x)dx\Rightarrow dy=\left( 2{{e}^{2x}} \right)dx
After applying the symbol of integration both sides:
dy=(2e2x)dx\Rightarrow \int{{}}dy=\int{\left( 2{{e}^{2x}} \right)}dx
After integrating the above equation, we will get:
y=2e2x2 y=e2x \begin{aligned} & \Rightarrow y=\dfrac{2{{e}^{2x}}}{2} \\\ & \Rightarrow y={{e}^{2x}} \\\ \end{aligned}
Now, we got the given equation of the question from the integration of the solution. Hence, the solution is correct.