# Math Help - Another differential.

1. ## Another differential.

$y = \sqrt {e^{5x}+3x}$

$= (e^{5x}+3x^2)^{\frac{1}{2}}$

$y'= \frac{1}{2} (e^{5x}+3x^2)^{\frac{-1}{2}} (5x+6x)$

Is that as far as i go , im not sure what to do now .

2. Originally Posted by el123
$y = \sqrt {e^{5x}+3x}$

$= (e^{5x}+3x^2)^{\frac{1}{2}}$

$y'= \frac{1}{2} (e^{5x}+3x^2)^{\frac{-1}{2}} (5x+6x)$

$y'= \frac{1}{2} (e^{5x}+3x^2)^{-\frac{1}{2}} (\textcolor{red}{5e^{5x}}+6x)$
correction

3. Originally Posted by el123
$y = \sqrt {e^{5x}+3x}$

$= (e^{5x}+3x^2)^{\frac{1}{2}}$

$y'= \frac{1}{2} (e^{5x}+3x^2)^{\frac{-1}{2}} (5e^{5x}+6x)$

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 $y'$, use Liebnitz notation. IE $\frac{dy}{dx}$

Then all you'd have too do is multiply both sides by $dx$. Like so

$dx\left(\frac{dy}{dx}\right)= \frac{1}{2} (e^{5x}+3x^2)^{\frac{-1}{2}} (5e^{5x}+6x)dx$

4. so if im asked to find the slope of that equation when x = 1 how do i do that?

Also could you please explain where the correction came from.

And i really appreciate all your help skeeter. You rock.

5. Originally Posted by el123
so if im asked to find the slope of that equation when x = 1 how do i do that?
Plug 1 into the derivative.

Originally Posted by el123
Also could you please explain where the correction came from.
Recal that $\frac{d}{dx}[e^u]=e^uu'$. in your problem $\frac{d}{dx}[e^{5x}]=e^{5x}\cdot{5}$

Originally Posted by el123
And i really appreciate all your help skeeter. You rock.
I rock too

6. Originally Posted by el123
so if im asked to find the slope of that equation when x = 1 how do i do that?

Also could you please explain where the correction came from.
the derivative of $e^{f(x)}$ is $e^{f(x)} \cdot f'(x)$

so, the derivative of $e^{5x}$ is $e^{5x} \cdot 5 = 5e^{5x}$

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.

7. ok now im confused , so i am being asked to 'differentiate'

Can you please explain where i begin to 'differentiate' as opposed to finding the derivative?

8. Originally Posted by el123
ok now im confused , so i am being asked to 'differentiate'

Can you please explain where i begin to 'differentiate' as opposed to finding the derivative?
Yeah, it's easy. The process of finding the derivative is called differentiation.

So, you differentiated $y$ to find the derivative $y'$.

I thought you wanted the differential in my first post, so forget that ever happened. You will know what differentials are soon enough...