Answer:
[tex]\frac{dy}{dx}=3x^{3x}(lnx+1)[/tex]
Step-by-step explanation:
We are given that a function
[tex]y=x^{3x}+1[/tex]
We have to find the derivative of the function
Let [tex]u=x^{3x}[/tex]
[tex]y=u+1[/tex]
Taking ln on both sides
[tex]lnu=3xln x[/tex]
By using [tex]lna^b=blna[/tex]
Differentiate w.r.t x
[tex]\frac{1}{u}\frac{du}{dx}=3(lnx+x\times \frac{1}{x})=3(lnx+1)[/tex]
[tex]\frac{d(lnx)}{dx}=\frac{1}{x}[/tex]
[tex]\frac{d(u\cdot v)}{dx}=u'v+v'u[/tex]
[tex]\frac{du}{dx}=3u(lnx+1)=3x^{3x}(lnx+1)[/tex]
Differentiate y w.r.t x
[tex]\frac{dy}{dx}=\frac{du}{dx}[/tex]
Using the value of du/dx
[tex]\frac{dy}{dx}=3x^{3x}(lnx+1)[/tex]
