quick question as I'm sure this is simple but it has me stumped.
I have to integrate and differentiate this equation. Not sure on the exponential, had a couple of goes but it doesn't look right.
Here is the equation:
$$\frac{1}{2}e^{x/2} +\frac{1}{2x}-\frac{3}{2}=0$$
Any help would be greatly appreciated :]
Okay so the integral can be evaluated to be
$$ \begin{align*} \int \frac{e^{x/2}}{2}+ \frac{1}{2x}− \frac{3}{2} \; dx = e^{x/2} + \frac{\ln{x}}{2} - \frac{3x}{2} + c = c \end{align*} $$
and the derivative
$$ \frac{d}{dx}\left(\frac{e^{x/2}}{2}+ \frac{1}{2x}− \frac{3}{2}\right) = \frac{e^{x/2}}{4} - \frac{1}{2x^2} = 0 $$
So, $$f(x) = \frac{e^\frac{x}{2}}{2} + \frac{1}{2x} - \frac{3}{2}$$ Rewrite this to make it easier to differentiate/integrate $$f(x) = \frac{1}{2}e^\frac{x}{2} + \frac{1}{2}x^{-1} - \frac{3}{2}$$
Differentiating we get $$f'(x) = \frac{1}{4}e^\frac{x}{2} - \frac{1}{2}x^{-2}$$
Integrating we get $$\int{f(x)dx} = \int{\frac{1}{2}e^\frac{x}{2} + \frac{1}{2}x^{-1} - \frac{3}{2} dx} $$
$$\int{f(x)dx} = e^{\frac{x}2} + \frac12\log{x} - \frac32x + C$$
Quick bit about exponentials:
To differentiate $f(x) = e^{Ax}$ we multiply $f(x)$ by the derivative of $Ax$. So, $$f'(x) = Ae^{Ax}$$
To integrate $f(x) = e^{Ax}$ we divide $f(x)$ by the derivative of $Ax$. So, $$\int{f(x)dx} = \frac{e^{Ax}}A$$
