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Question: How do you find the derivative of \[{{e}^{2x}}\] using the product rule?...

How do you find the derivative of e2x{{e}^{2x}} using the product rule?

Explanation

Solution

Assume the given function as y. Now, write the given function e2x{{e}^{2x}} as ex×ex{{e}^{x}}\times {{e}^{x}}. Consider y as the product of two exponential functions. Now, apply the product rule of differentiation given as: - d(u×v)dx=udvdx+vdudx\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}. Here, consider, u=exu={{e}^{x}} and v=exv={{e}^{x}}. Use the formula: - dexdx=ex\dfrac{d{{e}^{x}}}{dx}={{e}^{x}} to simplify the derivative and get the answer.

Complete step by step solution:
Here, we have been provided with the function e2x{{e}^{2x}} and we are asked to find its derivative using the product rule. Let us assume the given function as y. So, we have,
y=e2x\Rightarrow y={{e}^{2x}}
The above expression can be written as:
y=ex+x\Rightarrow y={{e}^{x+x}}
Using the formula: am+n=am×an{{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}, we get,
y=ex×ex\Rightarrow y={{e}^{x}}\times {{e}^{x}}
Now, we can assume the given functions as two exponential functions (ex{{e}^{x}}). So, we have. Let us assume one ex{{e}^{x}} as ‘u’ and the other as ‘v’ respectively. So, we have,
y=u×v\Rightarrow y=u\times v
Differentiating both sides with respect to x, we get,
dydx=d(u×v)dx\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( u\times v \right)}{dx}
Now, applying the product rule of differentiation given as: -
d(u×v)dx=udvdx+vdudx\Rightarrow \dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}, we get,
dydx=udvdx+vdudx\Rightarrow \dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}
Substituting the assumed values of u and v, we get,
dydx=exd(ex)dx+exd(ex)dx\Rightarrow \dfrac{dy}{dx}={{e}^{x}}\dfrac{d\left( {{e}^{x}} \right)}{dx}+{{e}^{x}}\dfrac{d\left( {{e}^{x}} \right)}{dx}
We know that: dexdx=ex\dfrac{d{{e}^{x}}}{dx}={{e}^{x}}, so we have,

& \Rightarrow \dfrac{dy}{dx}={{e}^{x}}\left( {{e}^{x}} \right)+{{e}^{x}}\left( {{e}^{x}} \right) \\\ & \Rightarrow \dfrac{dy}{dx}={{e}^{2x}}+{{e}^{2x}} \\\ & \Rightarrow \dfrac{dy}{dx}=2{{e}^{2x}} \\\ \end{aligned}$$ Hence, the above relation is our answer. **Note:** One may note that we can also use the chain rule of derivative to get the answer. What we will do is we will first differentiate the function $${{e}^{2x}}$$ with respect to 2x and then differentiate 2x with respect to x, in the next step we will consider their product. In this way also we will get the same answer. But here we were asked to use the product rule and that is why we did not apply the chain rule. You must remember all the basic rules and formulas of differentiation like: - product rule, chain rule, $$\dfrac{u}{v}$$ rule etc.