Question

How do you find the derivative of ${x^{2x}}$?

Answer 1

In this question, we have to find the derivative of the given quantity. Consider the given quantity equal to some variable and then take logarithm both the sides of the equation and use the property, $\log {a^b} = b\log a$. After that, differentiate it with respect to x using the product rule of the derivative. After that do simplification and substitute the value of $y$ to get the desired result.

Complete step-by-step answer:
Let $y = {x^{2x}}$ ….. (1)
Take log on both sides we have,
$\Rightarrow \log y = \log {x^{2x}}$
Now according to logarithmic property $\log {a^b} = b\log a$, we can write the above equation as
$\Rightarrow \log y = 2x\log x$
Now apply the chain rule of differentiation we have,
$\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}\left( b \right) + b\dfrac{d}{{dx}}\left( a \right)$
So, differentiate the above equation with respect to $x$ what we have,
$\Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) = 2x\dfrac{d}{{dx}}\left( {\log x} \right) + \left( {\log x} \right)\dfrac{d}{{dx}}\left( {2x} \right)$
Differentiate the terms,
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{{2x}}{x} + \left( {\log x} \right) \times 2$
Cancel out the common factor,
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = 2 + 2\log x$
Take 2 commons on the right side,
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = 2\left( {1 + \log x} \right)$
Multiplying both sides by $y$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 2y\left( {1 + \log x} \right)$
Substitute the value of $y$ from equation (1),
$\Rightarrow \dfrac{{dy}}{{dx}} = 2{x^{2x}}\left( {1 + \log x} \right)$

Hence, the derivative of ${x^{2x}}$ is $2{x^{2x}}\left( {1 + \log x} \right)$.

Note:
Whenever we face such types of problems the key concept is simply to make use of logarithm, as exponential powers need to be changed to a simpler form before starting doing the derivative part. A good understanding of some logarithm properties, product rule of derivatives along properties of logarithm helps to get on the right track to reach the answer.
Logarithms can make multiplication and division of large numbers easier, because adding logarithms is the same as multiplying, and subtracting logarithms is the same as dividing.
Change of base rule law,
${\log _y}x = \dfrac{{\log x}}{{\log y}}$
Product rule law,
$\log xy = \log x + \log y$
Quotient rule law,
$\log \dfrac{x}{y} = \log x - \log y$
Power rule law,
$\log {x^y} = y\log x$