Question

If y= ${x^x}$,then find $\dfrac{{dy}}{{dx}}$ .

Answer 1

Hint: In this question, first we take log on both sides. After this, we differentiate on both sides w.r.t. ‘x’. The term on LHS logy will give the $\dfrac{{dy}}{{dx}}$term and on RHS, we will use the product rule of differentiation. Finally, rearrange the terms to get the answer.

Complete step-by-step answer:
The given expression is:
y= ${x^x}$.
Since the term on RHS is exponential. So, we will take logs on both sides.
On taking log on both sides, we get:
log y = xlogx.
On differentiating on both sides w.r.t. ‘x’, we get:
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} + \log x$
On solving the terms on RHS, we get:
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = 1 + \log x$
On multiplying by ‘y’ on both sides, we get:
$\dfrac{{dy}}{{dx}} = (1 + \log x)y$
Putting the value of ‘y’ in above equation, we have:
$ \Rightarrow $ $\dfrac{{dy}}{{dx}} = (1 + \log x){x^x}$
Therefore the required answer is $\dfrac{{dy}}{{dx}} = (1 + \log x){x^x}$.

Note: In this type of question which involves exponent terms, and it is asked to find $\dfrac{{dy}}{{dx}}$, the rule is to take log on both sides and then proceed. You must remember some of the important rules of differentiation.
1.Product rule: used to find differentiation of two functions in multiplication.
$\dfrac{{d\left( {f(x)g(x)} \right)}}{{dx}} = \dfrac{{df(x)}}{{dx}}g(x) + \dfrac{{dg(x)}}{{dx}}f(x)$
2. chain rule: used to find differentiation of a composite function.
$\dfrac{{d(f(g(x))}}{{dx}} = f'(g(x))g'(x)$