Question

Find the value of $\dfrac{d}{{dx}}({x^x})$
A) ${x^x}\log (\dfrac{e}{x})$
B) ${x^x}\log ex$
C) $\log ex$
D) ${x^x}\log x$

Answer 1

Here we assume the given function as a variable. Apply log on both sides of the equation and open the right side using the property of logarithm. Differentiate both sides with respect to x keeping in mind the variable we assumed is also a function of x. Apply product rule of differentiation on RHS and chain rule of differentiation on LHS.

Formula used:
*$\log {m^n} = n\log m$
*$\log m + \log n = \log mn$

• Product rule of differentiation:$\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}(b) + b\dfrac{d}{{dx}}(a)$
• Chain rule of differentiation:$\dfrac{d}{{dx}}g\left( {f(x)} \right) = \dfrac{d}{{dx}}g(f(x)) \times \dfrac{d}{{dx}}f(x)$

Complete step-by-step answer:
We have to find$\dfrac{d}{{dx}}({x^x})$
Here the function is ${x^x}$
Let us assume the function equal to a variable y.
Let $y = {x^x}$ … (1)
Now we apply log function on both sides of the equation.
$\Rightarrow \log y = \log ({x^x})$
Use the property $\log {m^n} = n\log m$ where m is x and n is x.
$\Rightarrow \log y = x(\log x)$
Now differentiate both sides of the equation with respect to x
$\Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) = \dfrac{d}{{dx}}\left( {x(\log x)} \right)$
Apply chain rule of differentiation in LHs of the equation
Chain rule gives us$\dfrac{d}{{dx}}g\left( {f(x)} \right) = \dfrac{d}{{dx}}g(f(x)) \times \dfrac{d}{{dx}}f(x)$.
Here $g(f(x)) = \log (y),f(x) = y$, then the equation becomes
$\Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) \times \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {x(\log x)} \right)$
We know $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
$\Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {x(\log x)} \right)$
Now apply product rule of differentiation in RHS of the equation
Product rule gives us$\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}(b) + b\dfrac{d}{{dx}}(a)$
Here$a = x,b = \log x$, then the equation becomes
$\Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = x\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}x$
Substitute the values$\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$and$\dfrac{{dx}}{{dx}} = 1$in RHS of the equation
$\Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} + \log x \times 1$
Cancel same terms from numerator and denominator
$\Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = 1 + \log x$
Multiply both sides of the equation by y
$\Rightarrow y \times \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = y \times (1 + \log x)$
Cancel same function from numerator and denominator in LHS of the equation
$\Rightarrow \dfrac{{dy}}{{dx}} = y(1 + \log x)$
Substitute the value of $y = {x^x}$from equation (1)
$\Rightarrow \dfrac{d}{{dx}}({x^x}) = {x^x}(1 + \log x)$
Now we can write $\log e = 1$in RHs of the equation
$\Rightarrow \dfrac{d}{{dx}}({x^x}) = {x^x}(\log e + \log x)$
Use the property $\log m + \log n = \log mn$in RHs of the equation
$\Rightarrow \dfrac{d}{{dx}}({x^x}) = {x^x}(\log ex)$

So, the correct answer is “Option B”.

Note: Students might get stuck at the point where the answer comes out ${x^x}(1 + \log x)$, but we don’t have a similar option available to choose from. We can substitute the value of $\log e = 1$ as logarithm cancels out the exponential function giving the answer 1. Many students make the mistake of not substituting the value of the assumed variable back in the solution in the end.