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Question: How do you differentiate \(f\left( x \right)=\tan \left( {{e}^{x}} \right)\) using the chain rule?...

How do you differentiate f(x)=tan(ex)f\left( x \right)=\tan \left( {{e}^{x}} \right) using the chain rule?

Explanation

Solution

In this question we have a composite function which has no direct formula for calculating the derivative therefore, we will use the chain rule on the function which is f(x)=g(h(x))h(x)f'(x)=g'(h(x))h'(x).
We will consider in the function the outer function to be g(x)=tanxg(x)=\tan x and the inner function h(x)=exh(x)={{e}^{x}}. We will then differentiate the terms and simplify to get the required solution.

Complete step-by-step solution:
We have the equation given as:
f(x)=tan(ex)\Rightarrow f\left( x \right)=\tan \left( {{e}^{x}} \right)
Since we have to find the derivative of the term, it can be written as:
f(x)=ddxtan(ex)\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\tan \left( {{e}^{x}} \right)
Now since there is no direct formula for calculating the derivative of the given expression, we will use the chain rule which is f(x)=g(h(x))h(x)f'(x)=g'(h(x))h'(x).
We know that ddxtanx=sec2x\dfrac{d}{dx}\tan x={{\sec }^{2}}x, and since we are using chain rule, we will write it as:
sec2(ex)ddxex\Rightarrow {{\sec }^{2}}\left( {{e}^{x}} \right)\dfrac{d}{dx}{{e}^{x}}
We know that ddxex=ex\dfrac{d}{dx}{{e}^{x}}={{e}^{x}} therefore, on substituting, we get:
sec2(ex)×ex\Rightarrow {{\sec }^{2}}\left( {{e}^{x}} \right)\times {{e}^{x}}
On simplifying the term, we get:
sec2(ex)ex\Rightarrow {{\sec }^{2}}\left( {{e}^{x}} \right){{e}^{x}}, which is the required solution.

Note: It is to be remembered that the chain rule is a different rule than the product and the quotient rule. It is to be remembered that there can be more than two functions in a composite function. The general rule to differentiate is by first solving the outer function and then moving to the inward functions.
It is to be also remembered that differentiation is the inverse of integration. If the integration of a term aa is bb, then the differentiation of the term bb will be aa and vice versa.