Answer:
[tex]\dfrac{dy}{dx} = x^{2x}(2\ln{\left (x \right )} + 2)[/tex]
Step-by-step explanation:
Given
[tex]y = x^{2x}[/tex]
we first need to rewrite this in a form that we know we can differentiate! apply natural log on both sides
[tex]\ln{(y)} = \ln{(x^{2x)}[/tex]
[tex]\ln{(y)} = 2x\ln{(x)}[/tex]
[tex]y = e^{2x\ln{(x)}}[/tex]
it is to be noted that we two representations of y
[tex]y = e^{2x\ln{(x)}} = x^{2x}[/tex]
as we know that [tex]\frac{d}{dx}(e^{f(x)})=e^{f(x)}(f'(x))[/tex], we can use the same rule here.
by using the product rule we can differentiate 2xln(x)
[tex]\dfrac{dy}{dx} = e^{2x\ln{(x)}}\left(\dfrac{d}{dx}(2x\ln{(x)})\right)[/tex]
[tex]\dfrac{dy}{dx} = e^{2x\ln{(x)}}(2\ln{\left (x \right )} + 2)[/tex]
this is our answer and it can also be written as:
[tex]\dfrac{dy}{dx} = x^{2x}(2\ln{\left (x \right )} + 2)[/tex]
