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Question

Question: How would you find the derivative of \[y = \ln {(\ln x)^2}\] ?...

How would you find the derivative of y=ln(lnx)2y = \ln {(\ln x)^2} ?

Explanation

Solution

Here the question need to find the derivative of the given function with respect of the variable on the right hand side of the equation, here we are two derivable quantity on the other side which we have to derive and for the function it need to be solved together, for which first we have to find the derivative of first derivable quantity and then for the second derivable quantity.

Formulae Used:
Derivative of “lnx”
ddxlnx=1x\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}

Complete step by step solution:
The given question is y=ln(lnx)2y = \ln {(\ln x)^2}. Here we have to use the property of derivation for “lnx” to move further in the question and then solve, the property says:
ddxlnx=1x\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}
By using this formulae we can solve for the given derivation, on applying this to our question we get:

\Rightarrow \dfrac{{dy}}{{dx}}= \dfrac{1}{{{{(\ln x)}^2}}}\dfrac{d}{{dx}}{(\ln x)^2} \\\ \Rightarrow \dfrac{{dy}}{{dx}}= \dfrac{1}{{{{(\ln x)}^2}}}2\ln x \times \dfrac{d}{{dx}}\ln x \\\ \Rightarrow \dfrac{{dy}}{{dx}}=\dfrac{1}{{{{(\ln x)}^2}}}2\ln x \times \dfrac{1}{x} \\\ \therefore\dfrac{{dy}}{{dx}} = \dfrac{2}{{x \times \ln x}}$$ Here we have first found the derivation of outer “ln” and then for the square term of the inner “ln” and then after solving for the square we move toward the inner “ln” alone and finally no need to solve for “x” because its derivative will be one. **Hence, the derivative of $$y = \ln {(\ln x)^2}$$ is $\dfrac{2}{{x \times \ln x}}$.** **Note:** To find the derivative of any integrated function, first we have to solve the outermost function, and then moving forward we have to solve for every function or variable present inside, and finally all the derivative will be put in the answer in product form, and the obtained answer will be our required answer.