Question

How do you differentiate $f(x)={{e}^{\tan \left( \dfrac{1}{x} \right)}}$ using the chain rule?

Answer 1

Hint: The given expression is to be differentiated using the chain rule. We will first differentiate the exponential function which is nothing but exponential function itself. Then, we will differentiate the functions present in the power raised in the exponential function and that is $\tan \left( \dfrac{1}{x} \right)$. We will then proceed on to differentiate the angle of tangent function, that is, $\dfrac{1}{x}$. Hence, we differentiate the given expression.

Complete step-by-step solution:
According to the given question, we have an expression which we have to differentiate using the chain rule.
Chain rule refers to the sequence in which the function in a given expression is differentiated.
For example – if we have to differentiate \[f(x)=\sin {{x}^{2}}\]
Then we will first differentiate the sine function which will give us cosine function and then we will differentiate the \[{{x}^{2}}\] which will give us \[2x\]. So, we have,
\[f'(x)=\cos {{x}^{2}}.2x\]
The given expression we have is,
$f(x)={{e}^{\tan \left( \dfrac{1}{x} \right)}}$----(1)
First, we will differentiate the exponential term, that is we get,
\[f'(x)={{e}^{\tan \left( \dfrac{1}{x} \right)}}\] ----(2)
As we know that, \[\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}\].
Next, we will differentiate the \[\tan \left( \dfrac{1}{x} \right)\] which is the power raised to, in the exponential function.
In continuation, we get the equation (2) as,
\[f'(x)={{e}^{\tan \left( \dfrac{1}{x} \right)}}.{{\sec }^{2}}\left( \dfrac{1}{x} \right)\]-----(3)
As we know that, \[\dfrac{d}{dx}(\tan x)={{\sec }^{2}}x\].
Now, we will differentiate the \[\left( \dfrac{1}{x} \right)\] function and we get the differentiation continuation to equation (3), we have,
\[f'(x)={{e}^{\tan \left( \dfrac{1}{x} \right)}}.{{\sec }^{2}}\left( \dfrac{1}{x} \right).\left( -\dfrac{1}{{{x}^{2}}} \right)\]------(4)
As we know that, \[\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=-\dfrac{1}{{{x}^{2}}}\].
Therefore, we have the differentiation of the given expression as:
\[f'(x)={{e}^{\tan \left( \dfrac{1}{x} \right)}}.{{\sec }^{2}}\left( \dfrac{1}{x} \right).\left( -\dfrac{1}{{{x}^{2}}} \right)\]

Note: The differentiation of each function in the expression is important while using chain rule and should be done carefully. Also, a particular sequence is followed in chain rule and should be strictly adhered and not otherwise, that is, first the outer function is differentiated and we then proceed to the inner function, that is,
If we have to differentiate \[h(x)=f(g(x))\], we will get,
\[h'(x)=f'(g(x)).g'(x)\]