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Last updated date: 29th Feb 2024
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IVSAT 2024
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Hint: First express the given function in the form of $y = f\left( x \right)$. Then take logarithm on both sides of the equation to simplify it so as to make differentiation easier. Then differentiate both sides with respect to $x$ and make the substitution wherever required. Apply the chain rule of differentiation which is $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}}$ to differentiate the composite functions.

Complete step by step answer:
According to the question, we have to show how to differentiate the function ${x^{\ln x}}$.
Let this function be denoted as $y$. Then we have:
$ \Rightarrow y = {x^{\ln x}}{\text{ }}.....{\text{(1)}}$
If we take natural logarithm on both sides, we’ll get:
 \[ \Rightarrow \ln y = \ln {x^{\ln x}}\]
We know the logarithmic formula $\ln {a^b} = b\ln a$. On applying this formula, we have:
\[ \Rightarrow \ln y = \ln x\ln x\]
This can be written as:
\[ \Rightarrow \ln y = {\left( {\ln x} \right)^2}\]
Now if we put $\ln x = t$, we’ll get:
\[ \Rightarrow \ln y = {t^2}\]
Differentiating this equation both sides with respect to $x$, we’ll get:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\ln y} \right) = \dfrac{d}{{dx}}\left( {{t^2}} \right)\]
We know the chain rule of differentiation as shown below:
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}}$
Applying this rule for our function, we’ll get:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\ln y} \right) = \dfrac{d}{{dt}}\left( {{t^2}} \right)\dfrac{{dt}}{{dx}}\]
We know the formulas of differentiation $\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}$ and $\dfrac{d}{{dx}}{x^2} = 2x$. Applying them, we’ll get:
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = 2t\dfrac{{dt}}{{dx}}\]
Putting back the value of $t$, we’ll get:
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = 2\ln x\dfrac{d}{{dx}}\left( {\ln x} \right)\]
Again applying the formula of differentiation, we’ll get:
   \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = 2\ln x \times \dfrac{1}{x} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2y\ln x}}{x} \\
Putting back the value of $y$ from equation (1), we’ll get:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{x^{\ln x}}} \right) = \dfrac{{2{x^{\ln x}}\ln x}}{x}\]
This is the required differentiation.

Note: Whenever we have to differentiate a composite function, we always use chain rule of differentiation after substitution. This makes a complex looking function simple from where we can differentiate step by step. For example, consider the given composite function:
$ \Rightarrow y = f\left( {g\left( x \right)} \right)$
To differentiate this function, we’ll substitute $g\left( x \right) = t$, we will have:
$ \Rightarrow y = f\left( t \right)$
Now we can apply chain rule of differentiation as shown below:
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}f\left( t \right) \times \dfrac{{dt}}{{dx}}$
Now this differentiation is simple and we can do it step by step. After doing this, we can put back the value of $t$ to get the answer.
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