Question

Given $ y = {e^{{{(\ln x)}^2}}} $ how do you find $ y'(e) $ ?

Answer 1

Hint:Differentiation is said to be a process of dividing a whole quantity into very small ones, in the given question we have to differentiate y with respect to x. We will first differentiate the whole quantity $ {e^{{{(\ln x)}^2}}} $ and then differentiate the quantity in its power as it is also a function of x ( $ {(\ln x)^2} $ , but we see that the term in the power itself has a power 2 so we differentiate that term $ (\ln x) $ . The result of multiplying these three differentiating will give the value of $ \dfrac{{dy}}{{dx}} $ or $ y'(x) $ , to find the value of $ y'(e) $ we put the value of x as equal to e in the differentiated function. On solving, we will get the correct answer.

Complete step by step answer:
We are given that $ y = {e^{{{(\ln x)}^2}}} $ and we have to find $ y'(e) $ that is we have to differentiate the function y.
 $
y = {e^{{{(\ln x)}^2}}} \\
\dfrac{{dy}}{{dx}} = \dfrac{{d[{e^{{{(\ln x)}^2}}}]}}{{dx}} \\
 $
Now, we know that the derivative of an exponential function is the function itself, that is the function remains unchanged.
 $
\dfrac{{d({e^x})}}{{dx}} = {e^x} \\
\Rightarrow \dfrac{{d{e^{{{(\ln x)}^2}}}}}{{dx}} = {e^{{{(\ln x)}^2}}}\dfrac{{d{{(\ln x)}^2}}}{{dx}}
\\
 $
We know, $ \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}} $
 $
\Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\ln }^2}x}}2{(\ln x)^{2 - 1}}\dfrac{{d\ln x}}{{dx}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\ln }^2}x}}2\ln x\dfrac{{d\ln x}}{{dx}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\ln }^2}x}}2\ln x(\dfrac{1}{x}) \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2{e^{{{\ln }^2}x}}\ln x}}{x} \\

 $
So, $ y'(x) = \dfrac{{2{e^{{{\ln }^2}x}}\ln x}}{x} $
Therefore, $ y'(e) = \dfrac{{2{e^{{{\ln }^2}e}}\ln e}}{e} $
Now,
 $
\ln e = 1 \\
\Rightarrow {\ln ^2}e = 1 \\
 $
 $
y'(e) = \dfrac{{2{e^1}(1)}}{e} = \dfrac{{2e}}{e} \\
\Rightarrow y'(e) = 2 \\
 $

Hence $ y'(e) = 2 $

Note: Usually, the rate of change of something is observed over a specific duration of time, but if we have to find the instantaneous rate of change of a quantity then we differentiate it, in the expression $ \dfrac{{dy}}{{dx}} $ , $ dy $ represents a very small change in the quantity and $ dx $ represents the small change in the quantity with respect to which the given quantity is changing. In the given question, we have a function of x, so by putting different values of x, we can find the instantaneous change in x at that particular value.