$\displaystyle y = \sqrt {e^{5x}+3x} $
$\displaystyle = (e^{5x}+3x^2)^{\frac{1}{2}}$
$\displaystyle y'= \frac{1}{2} (e^{5x}+3x^2)^{\frac{-1}{2}} (5x+6x)$
Is that as far as i go , im not sure what to do now .
If it is the differential that you want, first start by changing your notation. Instead of $\displaystyle y'$, use Liebnitz notation. IE $\displaystyle \frac{dy}{dx}$
Then all you'd have too do is multiply both sides by $\displaystyle dx$. Like so
$\displaystyle dx\left(\frac{dy}{dx}\right)= \frac{1}{2} (e^{5x}+3x^2)^{\frac{-1}{2}} (5e^{5x}+6x)dx$
the derivative of $\displaystyle e^{f(x)}$ is $\displaystyle e^{f(x)} \cdot f'(x)$
so, the derivative of $\displaystyle e^{5x}$ is $\displaystyle e^{5x} \cdot 5 = 5e^{5x}$
to find the value of the derivative at x = 1, plug in 1 for x.
btw ... what you are finding is a derivative, not a differential.
Yeah, it's easy. The process of finding the derivative is called differentiation.
So, you differentiated $\displaystyle y$ to find the derivative $\displaystyle y'$.
I thought you wanted the differential in my first post, so forget that ever happened. You will know what differentials are soon enough...