Question

Differentiate Functions of Other Bases In Exercise, find the derivative of the function.

Answer 1

Answer:

$\frac{dy}{dx}=3x^{3x}(lnx+1)$

Step-by-step explanation:

We are given that a function

$y=x^{3x}+1$

We have to find the derivative of the function

Let $u=x^{3x}$

$y=u+1$

Taking ln on both sides

$lnu=3xln x$

By using $lna^b=blna$

Differentiate w.r.t x

$\frac{1}{u}\frac{du}{dx}=3(lnx+x\times \frac{1}{x})=3(lnx+1)$

$\frac{d(lnx)}{dx}=\frac{1}{x}$

$\frac{d(u\cdot v)}{dx}=u'v+v'u$

$\frac{du}{dx}=3u(lnx+1)=3x^{3x}(lnx+1)$

Differentiate  y w.r.t x

$\frac{dy}{dx}=\frac{du}{dx}$

Using the value of du/dx

$\frac{dy}{dx}=3x^{3x}(lnx+1)$