Question

How do you differentiate ${e^{{{\left( {\ln 2x} \right)}^2}}}$ with respect to x using the chain rule of differentiation?

Answer 1

Hint: In the given problem, we are required to differentiate ${e^{{{\left( {\ln 2x} \right)}^2}}}$ with respect to x. Since, ${e^{{{\left( {\ln 2x} \right)}^2}}}$ is a composite function, so we will have to apply chain rule of differentiation in the process of differentiating ${e^{{{\left( {\ln 2x} \right)}^2}}}$ . So, differentiation of ${e^{{{\left( {\ln 2x} \right)}^2}}}$ with respect to x will be done layer by layer using the chain rule of differentiation. Also derivatives of ${e^x}$ and $\ln x$ with respect to $x$ must be remembered in order to tackle the given problem.

Complete step by step solution:
To find derivative of ${e^{{{\left( {\ln 2x} \right)}^2}}}$ with respect to x, we have to find differentiate ${e^{{{\left( {\ln 2x} \right)}^2}}}$with respect to x.
Let us assume the function $y = {e^{{{\left( {\ln 2x} \right)}^2}}}$
So, Derivative of ${e^{{{\left( {\ln 2x} \right)}^2}}}$ with respect to x can be calculated as \[\dfrac{d}{{dx}}\left( {{e^{{{\left( {\ln 2x} \right)}^2}}}} \right)\] .
Now, \[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{e^{{{\left( {\ln 2x} \right)}^2}}}} \right)\]
Now, we have to use the chain rule of differentiation for finding the derivative of the composite function ${e^{{{\left( {\ln 2x} \right)}^2}}}$.
According to chain rule, we can differentiate a composite function as
 \[\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = \dfrac{d}{{d\left[ {g\left( x \right)} \right] }}\left[ {f\left( {g\left( x \right)} \right)} \right] \times \dfrac{{d\left[ {g\left( x \right)} \right] }}{{dx}}\]
Now, Let us assume $u = {\left( {\ln 2x} \right)^2}$. Then, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{e^u}} \right)\]
Now, we know that derivative of ${e^x}$ with respect to x is ${e^x}$.So, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^u}\left( {\dfrac{{du}}{{dx}}} \right)\]
Now, substituting the value of $u$ back into the expression as ${\left( {\ln 2x} \right)^2}$, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\left( {\ln 2x} \right)}^2}}}\left( {\dfrac{{d{{\left( {\ln 2x} \right)}^2}}}{{dx}}} \right)\]
Now, let $t = \ln 2x$. So, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\left( {\ln 2x} \right)}^2}}}\left( {\dfrac{{d\left( {{t^2}} \right)}}{{dx}}} \right)\]
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\left( {\ln 2x} \right)}^2}}}\left( {2t} \right)\left( {\dfrac{{dt}}{{dx}}} \right)\]
Now, putting back $t$ as $\ln 2x$, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\left( {\ln 2x} \right)}^2}}}\left( {2\ln \left( {2x} \right)} \right)\left( {\dfrac{d}{{dx}}\left( {\ln \left( {2x} \right)} \right)} \right)\] because \[\dfrac{{dt}}{{dx}} = \dfrac{{d\left( {\ln \left( {2x} \right)} \right)}}{{dx}}\]
Now, let $r = 2x$. Then, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\left( {\ln 2x} \right)}^2}}}\left( {2\ln \left( {2x} \right)} \right)\left( {\dfrac{d}{{dx}}\left( {\ln \left( r \right)} \right)} \right)\]
Now, we know that the derivative of $\ln x$with respect to $x$is \[\left( {\dfrac{1}{x}} \right)\] . So, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\left( {\ln 2x} \right)}^2}}}\left( {2\ln \left( {2x} \right)} \right)\left( {\dfrac{1}{r}} \right)\dfrac{{dr}}{{dx}}\]
Now, putting back r as $2x$, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\left( {\ln 2x} \right)}^2}}}\left( {2\ln \left( {2x} \right)} \right)\left( {\dfrac{1}{{2x}}} \right)\dfrac{{d\left( {2x} \right)}}{{dx}}\]
Now, according to the power rule of differentiation, we have the derivative of ${x^n}$ as $n{x^{n - 1}}$. So, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{{{\left( {\ln 2x} \right)}^2}}}\left( {2\ln \left( {2x} \right)} \right)\left( {\dfrac{1}{{2x}}} \right)\left( 2 \right)\]
Now, simplifying the expression, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2{e^{{{\left( {\ln 2x} \right)}^2}}}\ln \left( {2x} \right)}}{x}\]
So, the derivative of ${e^{{{\left( {\ln 2x} \right)}^2}}}$ with respect to $x$is \[\dfrac{{2{e^{{{\left( {\ln 2x} \right)}^2}}}\ln \left( {2x} \right)}}{x}\] .
So, the correct answer is “ \[\dfrac{{2{e^{{{\left( {\ln 2x} \right)}^2}}}\ln \left( {2x} \right)}}{x}\] ”.

Note: The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.