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Question

Question: How do you differentiate \(f(x) = \tan (\ln x)\) using the chain rule?...

How do you differentiate f(x)=tan(lnx)f(x) = \tan (\ln x) using the chain rule?

Explanation

Solution

This question is from the topic of differentiation. In this question we need to find the derivative of function f(x)=tan(lnx)f(x) = \tan (\ln x) using the chain rule. To solve this question we need to know the conditions of chain rule of differentiation and the derivative of functions tanx\tan x and lnx\ln x.

Complete step by step answer:
Let us try to solve this question in which we are asked to find the derivative of function f(x)=tan(lnx)f(x) = \tan (\ln x). Before differentiating this, let’s have a look at definition of chain rule, suppose a function f(x)=g(h(x))f(x) = g(h(x)) such that both gg and hh are differentiable with respect to xx then ff is also differentiable and its differentiation is given by f(x)=g(h(x))h(x)f'(x) = g'(h(x)) \cdot h'(x) where f(x)=d(f(x))dxf'(x) = \dfrac{{d(f(x))}}{{dx}} and similarly gg' and hh' are derivatives of functions gg and hhrespectively.
Now, let’s find the derivative of functionf(x)=tan(lnx)f(x) = \tan (\ln x). Function to derive f(x)=tan(lnx)f(x) = \tan (\ln x) is composition of differentiable functions ln(x)\ln (x) andtanx\tan x. So for the derivative of the function f(x)=tan(lnx)f(x) = \tan (\ln x) we can use the chain rule of differentiate. After applying chain rule to function f(x)=tan(lnx)f(x) = \tan (\ln x), we get
d(tan(lnx))dx=d(tan(lnx))dxd(lnx)dx\dfrac{{d(\tan (\ln x))}}{{dx}} = \dfrac{{d(\tan (\ln x))}}{{dx}} \cdot \dfrac{{d(\ln x)}}{{dx}} eq(1)eq(1)
As we know that d(tanx)dx=sec2x\dfrac{{d(\tan x)}}{{dx}} = {\sec ^2}x. So we have,
d(tan(lnx))dx=sec2(lnx)\dfrac{{d(\tan (\ln x))}}{{dx}} = {\sec ^2}(\ln x) eq(2)eq(2)
And, also we know that d(lnx)dx=1x\dfrac{{d(\ln x)}}{{dx}} = \dfrac{1}{x}. So we have,
d(lnx)dx=1x\dfrac{{d(\ln x)}}{{dx}} = \dfrac{1}{x} eq(3)eq(3)
Now, putting back the value of eq(2)eq(2) and eq(3)eq(3)ineq(1)eq(1), we get the derivative ofln(5x)\ln (5x).Hence the derivative of function
d(tan(lnx))dx=sec2(lnx)1x d(tan(lnx))dx=sec2(lnx)x\dfrac{{d(\tan (\ln x))}}{{dx}} = {\sec ^2}(\ln x) \cdot \dfrac{1}{x} \\\ \therefore\dfrac{{d(\tan (\ln x))}}{{dx}}= \dfrac{{{{\sec }^2}(\ln x)}}{x}

Note: To solve these types of questions in which we are asked to find the derivative of a given function. For solving this type of question we are required to have knowledge of how to find derivatives of a function, differentiability of common function and properties of differentiation such sum rule, product rule, division rule and chain rule.