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# The derivative of $y = {x^{{2^x}}}$ with respect to $x$ is:A. ${x^{{2^x}}}{2^x}\left( {\dfrac{1}{x} + \ln x\ln 2} \right)$B. ${x^{{2^x}}}\left( {\dfrac{1}{x}\ln x\ln 2} \right)$C. ${x^{{2^x}}}{2^x}\left( {\dfrac{1}{x}\ln x} \right)$D. ${x^{{2^x}}}{2^x}\left( {\dfrac{1}{x} + \dfrac{{\ln x}}{{\ln 2}}} \right)$

Last updated date: 09th Sep 2024
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Hint: We will first take logs on both sides and then simplify the equation. Then, differentiate the equation using product rule of derivative and formulas of derivative such as, $\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x}$ and $\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\ln a$. At last substitute the value of $y$ from the given equation.

Whenever we have an expression with power as $x$ and we have to find its derivative, we will first take ln of both sides.
On taking ln both of equation $y = {x^{{2^x}}}$ , we get,
$\ln y = \ln {x^{{2^x}}}$
Now, simplify the equation using the properties of log.
As, we know, $\ln \left( {{a^m}} \right) = m\ln a$, thus, we can write $\ln y = \ln {x^{{2^x}}}$ as,
$\ln y = {2^x}\ln x$
Now, differentiate both sides with respect to $x$, using the formulas of derivatives such as,
$\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x}$, $\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\ln a$
We will use product rule, $\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = \left( {\dfrac{d}{{dx}}f\left( x \right)} \right)g\left( x \right) + \left( {\dfrac{d}{{dx}}g\left( x \right)} \right)f\left( x \right)$ to find its derivative.
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = {2^x}\ln 2\left( {\ln x} \right) + \dfrac{{{2^x}}}{x} \\ \dfrac{1}{y}\dfrac{{dy}}{{dx}} = {2^x}\left( {\ln 2\left( {\ln x} \right) + \dfrac{1}{x}} \right) \\ \dfrac{{dy}}{{dx}} = y{2^x}\left( {\ln 2\left( {\ln x} \right) + \dfrac{1}{x}} \right) \\$
Substitute the value of $y$ from the given equation.
$\dfrac{{dy}}{{dx}} = {x^{{2^x}}}{2^x}\left( {\ln 2\left( {\ln x} \right) + \dfrac{1}{x}} \right)$
Hence, option A is correct.

Note: Properties of log used in this question is $\ln \left( {{a^m}} \right) = m\ln a$. The product rule of derivative states that, $\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = \left( {\dfrac{d}{{dx}}f\left( x \right)} \right)g\left( x \right) + \left( {\dfrac{d}{{dx}}g\left( x \right)} \right)f\left( x \right)$. The derivative of $\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x}$ and $\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\ln a$.