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Question: The derivative of \(y = {x^{{2^x}}}\) with respect to \(x\) is: A. \({x^{{2^x}}}{2^x}\left( {\dfra...

The derivative of y=x2xy = {x^{{2^x}}} with respect to xx is:
A. x2x2x(1x+ln⁡xln⁡2){x^{{2^x}}}{2^x}\left( {\dfrac{1}{x} + \ln x\ln 2} \right)
B. x2x(1xln⁡xln⁡2){x^{{2^x}}}\left( {\dfrac{1}{x}\ln x\ln 2} \right)
C. x2x2x(1xln⁡x){x^{{2^x}}}{2^x}\left( {\dfrac{1}{x}\ln x} \right)
D. x2x2x(1x+ln⁡xln⁡2){x^{{2^x}}}{2^x}\left( {\dfrac{1}{x} + \dfrac{{\ln x}}{{\ln 2}}} \right)

Explanation

Solution

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, ddx(ln⁡x)=1x\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x} and ddx(ax)=axln⁡a\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\ln a. At last substitute the value of yy from the given equation.

Complete step by step answer:
Whenever we have an expression with power as xx and we have to find its derivative, we will first take ln of both sides.
On taking ln both of equation y=x2xy = {x^{{2^x}}} , we get,
ln⁡y=ln⁡x2x\ln y = \ln {x^{{2^x}}}
Now, simplify the equation using the properties of log.
As, we know, ln⁡(am)=mln⁡a\ln \left( {{a^m}} \right) = m\ln a, thus, we can write ln⁡y=ln⁡x2x\ln y = \ln {x^{{2^x}}} as,
ln⁡y=2xln⁡x\ln y = {2^x}\ln x
Now, differentiate both sides with respect to xx, using the formulas of derivatives such as,
ddx(ln⁡x)=1x\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x}, ddx(ax)=axln⁡a\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\ln a
We will use product rule, ddx(f(x)g(x))=(ddxf(x))g(x)+(ddxg(x))f(x)\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.
1ydydx=2xln⁡2(ln⁡x)+2xx 1ydydx=2x(ln⁡2(ln⁡x)+1x) dydx=y2x(ln⁡2(ln⁡x)+1x)  \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 yy from the given equation.
dydx=x2x2x(ln⁡2(ln⁡x)+1x)\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⁡(am)=mln⁡a\ln \left( {{a^m}} \right) = m\ln a. The product rule of derivative states that, ddx(f(x)g(x))=(ddxf(x))g(x)+(ddxg(x))f(x)\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 ddx(ln⁡x)=1x\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x} and ddx(ax)=axln⁡a\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\ln a.