Question

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

Answer 1

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.