Derivatives: If $f(x)= 1/e^x $ then If  $f(x)= 1/e^x $ then $ƒ′(x) = ?$ A: $1/e^x⋅ln(e^x)$
If $ƒ′(x) = e^x$ then $ƒ(x) = ?$A: $x$
Are my solutions correct? 
A: Hint: $\dfrac{\operatorname d(\mathsf e^x)^{-1}}{\operatorname d x~~~~~~~~} ~=~ \dfrac{\operatorname d (\mathsf e^x)^{-1}}{\operatorname d (\mathsf e^x)~~~~}\cdot \dfrac{\operatorname d \mathsf e^x}{\operatorname d x~}$ by the chain rule.
A: $f(x) = \frac{1}{e^x} = e^{-x}$
Via the Chain rule, $f'(x) = (e^{-x})\cdot\frac{d}{dx}(-x)$ $= -e^{-x}$
A: First, notice that $\frac{1}{e^x}=e^{-x}$, so in other words using the chain rule, and substituting following into the chain rule; $f(x)=e^x$, $f'(x)=e^x$, $g(x)=-x$, $g'(x)=-1$; we get:
$$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$$
$$\frac{d}{dx}e^{-x}=e^{-x}(-1)=-e^{-x}$$
For the second one, derivative of $e^x$ as well as antiderivative of it is always the same thing (that's why $e^x$ is awesome in my opinion). So the answer will be $e^x+C$, where C is any constant. 
The $C$ is added due to fact, that when deriving a function and it has a constant term, you just get rid of it. Also $\frac{d}{dx}n$, where $n$ is any constant is always $0$.
