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Question

Question: If \[y = {e^{2x}}\], find the value of \(\dfrac{{dy}}{{dx}}\)....

If y=e2xy = {e^{2x}}, find the value of dydx\dfrac{{dy}}{{dx}}.

Explanation

Solution

We will first write the formula for differentiation of the exponential function. Then, we will also require to note down the chain rule of differentiation. Apply those of the given question and thus we have the answer.

Complete step-by-step answer:
We know that we have: ddx(ex)=ex\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x} and ddx(xn)=nxn1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}
And the chain rule is given by dydx=dydu.dudx\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}}.\dfrac{{du}}{{dx}}.
Now, let y=e2xy = {e^{2x}} and .u=2xu = 2x
Substituting u in y, we will get:-
y=eu\Rightarrow y = {e^u}
Now, let us find dydx\dfrac{{dy}}{{dx}} using the chain rule.dydx=eu×2 \Rightarrow \dfrac{{dy}}{{dx}} = {e^u} \times 2
dydx=dydu.dudx\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}}.\dfrac{{du}}{{dx}}
Substituting the values in the above mentioned formula, we will then obtain:-
dydx=ddu(eu).ddx(2x)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{du}}\left( {{e^u}} \right).\dfrac{d}{{dx}}(2x)
Now, using ddx(ex)=ex\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x} and ddx(xn)=nxn1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}, we will get:-
(Because here n = 1)
Now, substituting back the value of u in above expression, we will then get:-
dydx=2e2x\Rightarrow \dfrac{{dy}}{{dx}} = 2{e^{2x}}.

\therefore The required answer is 2e2x2{e^{2x}}.

Note: The common mistake made by students is that they forget to multiply the derivative of exponential function by the derivative of 2x which is 2. You must remember to apply the chain formula otherwise the solution will be wrong. Exponential function is very crucial in real life as well. It is used to show the exponential change in anything for instance population growth, bacteria growth on anything. The Euler’s number is also related to exponential function as well which has many applications in real life as well.