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Question

Question: How do you find the derivative of \(\ln (\ln x)\)?...

How do you find the derivative of ln(lnx)\ln (\ln x)?

Explanation

Solution

This sum is a bit complicated as it involves 22 formulae. The first one is the derivative of a logarithmic function and the secondly using the Chain rule. Chain Rule states that the derivative of f(g(x))=f(g(x)×g(x)f(g(x)) = {f^{'}}(g(x) \times {g^{'}}(x). It helps us to differentiate composite functions. The sum might go wrong if the student doesn’t apply the chain ruler chain rule properly. For the chain rule, the student should first apply the derivative to the given constant and then again apply the derivative to the remaining composite function.

Complete step by step solution:
Applying the chain rule which states that
ddxf[g(x)]=f(g(x))×g(x)\dfrac{d}{{dx}}f[g(x)] = f'(g(x)) \times {g^{'}}(x). Comparing it with the given sum, we can say that the ff is an external ln\ln while ggis the internal lnx\ln x.
Also the derivative of the logarithm is
ddx(lnx)=1x\dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x}
First step of the chain rule would be as follows:
ddxln(lnx)=1lnxddxlnx........(1)\dfrac{d}{{dx}}\ln (\ln x) = \dfrac{1}{{\ln x}}\dfrac{d}{{dx}}\ln x........(1)
Applying the derivative again to the second composite function
ddxln(lnx)=1lnx×1x........(2)\dfrac{d}{{dx}}\ln (\ln x) = \dfrac{1}{{\ln x}} \times \dfrac{1}{x}........(2)

Thus the derivative of ln(lnx)\ln (\ln x) is 1lnx×1x\dfrac{1}{{\ln x}} \times \dfrac{1}{x}

Note:
In this particular sum, the student should apply the chain rule properly. He/she should not apply the derivative of g(x)g(x) followed by f(x)f(x). It is important to learn all the formulae for derivatives as the numerical based on derivatives cannot be solved until and unless the student is aware of the formula. He/she should memorize the formulae thoroughly in order to solve the sum correctly.