Solve f^prime(x)=e^x+ae^-x(-x)

$$f ^ { \prime } ( x ) = e ^ { x } + a e ^ { - x } ( - x )$$

Solve for a

$$a=\frac{e^{2x}}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$e^{x}+ae^{-x}\left(-x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)x$$

Subtract e^{x} from both sides.

$$ae^{-x}\left(-x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)x-e^{x}$$

Reorder the terms.

$$-axe^{-x}=x\frac{\mathrm{d}}{\mathrm{d}x}(f)-e^{x}$$

The equation is in standard form.

$$\left(-\frac{x}{e^{x}}\right)a=-e^{x}$$

Divide both sides by -xe^{-x}.

$$\frac{\left(-\frac{x}{e^{x}}\right)a}{-\frac{x}{e^{x}}}=-\frac{e^{x}}{-\frac{x}{e^{x}}}$$

Dividing by -xe^{-x} undoes the multiplication by -xe^{-x}.

$$a=-\frac{e^{x}}{-\frac{x}{e^{x}}}$$

Divide -e^{x} by -xe^{-x}.

$$a=\frac{e^{2x}}{x}$$

Solve for f

$$f\in \mathrm{R}$$

$$a=\frac{e^{2x}}{x}\text{ and }x\neq 0$$