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Question: What is the derivative of \(\ln {{e}^{2x}}\)?...

What is the derivative of ln⁡e2x\ln {{e}^{2x}}?

Explanation

Solution

We have to apply the chain rule as the function is a composite function.
A composite function is of the form f(g(x))f\left( g\left( x \right) \right) and we write its derivative as f′(g(x))+g′(x)f'\left( g\left( x \right) \right)+g'\left( x \right)
Here we can consider the composite function of ln⁡,ex,2x\ln ,{{e}^{x}},2x.

Complete step by step answer:
Given,
Function is ln⁡e2x\ln {{e}^{2x}}
Determine the derivative of given function
Consider the function as:
F=ln⁡e2xF=\ln {{e}^{2x}} …(1)
Now we apply chain rule to equation (1) we get:
F′=d(ln⁡e2x)dxF'=\dfrac{d\left( \ln {{e}^{2x}} \right)}{dx} …(2)
Now split the terms as listed above by using the chain rule:
We know that the derivative of ln⁡x=1x\ln x=\dfrac{1}{x}.
The derivative of the function ex=ex{{e}^{x}}={{e}^{x}} and we have the derivation of
xn=nxn−1{{x}^{n}}=n{{x}^{n-1}} .
Here we have applied the chain rule which involves three function this can be understood by the following equation:
f(g(h(x)))=f′(g(h(x)))×g′(h(x))×h′(x)f\left( g\left( h\left( x \right) \right) \right)=f'\left( g\left( h\left( x \right) \right) \right)\times g'\left( h\left( x \right) \right) \times h'\left( x \right)
And we have considered the function as follows:
f(x)=ln⁡x g(x)=ex h(x)=2x \begin{aligned} & f\left( x \right)=\ln x \\\ & g\left( x \right)={{e}^{x}} \\\ & h\left( x \right)=2x \\\ \end{aligned}
Now using all this standard derivation in equation (1) we get:
F′=1e2x×e2x×2 ⇒F′=2 \begin{aligned} & F'=\dfrac{1}{{{e}^{2x}}}\times {{e}^{2x}}\times 2 \\\ & \Rightarrow F'=2 \\\ \end{aligned}

Note: There is another method to solve this question as the chain rule method can be quite confusing where more than two terms are involved therefore to make it easier we can use the substitution method:
Let us assume that e2x=u,dudx=2e2x{{e}^{2x}}=u,\dfrac{du}{dx}=2{{e}^{2x}}
Then we can write the function as:
F=ln⁡(u)F=\ln \left( u \right) …(1)
Now since we know that the derivative of ln⁡x=1x\ln x=\dfrac{1}{x}.
Therefore, taking derivation of equation (1) we get:
F′=1ududxF'=\dfrac{1}{u}\dfrac{du}{dx} …(2)
Now we can substitute the value of uu and dudx\dfrac{du}{dx} in equation (2) we get:
F′=1e2x×2e2x ⇒F′=2 \begin{aligned} & F'=\dfrac{1}{{{e}^{2x}}}\times 2{{e}^{2x}} \\\ & \Rightarrow F'=2 \\\ \end{aligned}
Hence, we see that the value obtained in either case is the same so any method which is found easier could be used to solve this question.
If you use the chain rule method you should be careful enough to consider each function involved in it.