If the function $f(x)=(x^3)/(e^{x/2})$ and $g(x)$ is the inverse function, how to find derivative of $g(x)$ at $x=1$? If the function $f(x)=(x^3)/(e^{x/2})$ and g(x) is the inverse function, how to find derivative of g(x) at x=1 ? 
Im finding it really difficult to find g(x).Help please! And is there any indirect way to find derivative of g(x) without finding g(x)?
 A: Let $f(x)$ be a function such that $\forall x\in  (R\equiv (-\infty,6]) \; \exists \;f(x)\in D$ , where $R$ is domain of the function, $D$ is the range of function.
https://www.desmos.com/calculator/13qav60399
As from the graph of the function, it is clear that it is one-one and onto on it's range and domain. 
Now there exists a function $g(y): y \in D \to x \in R $ i.e., $\forall y \in D\; \exists \;x \in R$, Such that $g(y)$ is inverse of $f(x)$.
Then $\exists \;\;$$g(f(x))=x$ which is an identity in $x \in R$. Intuitively, this function takes an $x\in R$ send it to a $f(x)\in D$ and then the function $g(y)$ takes an $f(x)\in D$ and sends it back to $x \in R$.
Now that we have proved the existence of inverse and found the identity function in x,  it's time to use chain rule to differentiate.

$$g'(\color{red}{f(x)})\times f'(\color{blue}{x})=1$$
$$\implies g'(\color{red}{f(x)})=\frac{1}{f'(\color{blue}{x})}$$
First let us find what is $\large{\frac{1}{f'(x)}}$, given $f(x)=\large{\frac{x^3}{e^{x/2}}}$
$$f'(x)=\frac{e^{x/2}\times 3x^2-\frac{1}{2}x^3\times e^{x/2}}{e^x}$$
$$f'(x)=\frac{6x^3-x^2}{2e^{x/2}}$$ 
$$\frac{1}{f'(\color{blue}{x})}=\frac{2\color{green}{e^{x/2}}}{\color{brown}{6x^2-x^3}}=g'(\color{red}{f(x)})$$
Now our only problem is to find an $x$ such that $\color{red}{f(x)}=1$ i.e., $\color{blue}{x}^3=e^{\color{blue}{x}/2}$


Here I am going to roughly introduce Lambert W function. 
  Consider a equation $Xe^{X}=Y$ then it can be written in terms of W function as $X=W(Y)$

Now in our equation, taking cube root both side.
$$\color{blue}{x}=e^{\color{blue}{x}/6}$$
Which can be rewritten as, 
$$\color{blue}{x}e^{-\color{blue}{x}/6}=1$$
Substitute $-\frac{\color{blue}{x}}{6}=t$ so that $\color{blue}{x}=-6t$.
$$-6te^t=1$$
$$te^t=\frac{-1}{6}$$
Solution to this is W function given by , 
$$t=W(-\frac{1}{6})$$ and now substituting into $\color{blue}{x}$,
$$\color{blue}{x}=-6W(-\frac{1}{6})$$

Now , we will find some coloured terms individually in our derivative expression of $g'(\color{red}{f(x)})$, now that we have a value of $\color{blue}{x}$.
First of all what is, $$\color{green}{e^{x/2}}=e^{-3W(-1/6)}$$

Here we will use an identity, $$e^{n.W(x)}=\left(\frac{x}{W(x)}\right)^n$$ shorthand notation $(W(x))^3=W^{3}(x)$

$$e^{-3W\left(-\frac16\right)}=\left(\frac{-\frac{1}{6}}{W\left(-\frac{1}{6}\right)}\right)^{-3}=-216W^{3}\left(-\frac{1}{6}\right)$$   

Secondly what is $ \color{brown}{6x^2-x^3}$
This is pretty straightforward, $$\color{brown}{6x^2-x^3}=216W^2(-\frac16)+216W^3(-\frac16)$$

Putting it all together, 
$$g'(\color{red}{1})=2\times \frac{\color{green}{-216W^3\left(-\frac{1}{6}\right)}}{\color{brown}{216W^2\left(-\frac16\right)+216W^3\left(-\frac16\right)}}$$
Which finally gives us, 
$$\bbox[black]{\color{white}{g'(1)=-2 \times \frac{W\left(-\frac{1}{6}\right)}{1+W\left(-\frac{1}{6}\right)}\approx 0.514083..}}$$ 
http://www.wolframalpha.com/input/?i=-2%281-1%2F%281%2BW%28-1%2F6%29%29%29
PS, I know that the black box looks ugly, but i have always wanted to use that. :)
A: When it exists (that is, when there is an inverse function), if you set $y=g(x)$, the general formula is:
$$g'(x)=\frac1{f'(y)}$$
provided $f'(y)\neq 0$. If $f'(y_0)=0$, $g$ is not differentiable at $x_0$.
The main problem consists in expressing the result as an explicit function of $x$. This cannot always be realised.
