How to use the chain rule to differentiate I have two examples of problems that I don't know how to differentiate.
y = $e^{x^2/3x+2}$      and     y = $-10x^{3x^2-4}$
I know to take the ln on both sides. I just don't understand whereto go afterwards. I am needing a good expanation so I can do these types of problems.
Edit: I don't know If i need to do these using logarithmic differentiation. Could someone explain the chain rule process applied to one of these?
 A: here is how you can do logarithmic differentiation.  let us take the second one. you can try the first one yourself.
$$|y| = 10 x^{3x^2 - 4}$$ and take the $\ln$ both side to get $$\ln|y| = \ln 10 + (3x^2 -4)\ln x$$ now differencing gives you $$\frac{dy}{y} = 6x\ln x\, dx + \frac{(3x^2-4)}x \,dx$$ you can turn that into $$\frac{dy}{dx} = \frac{y(6x^2\ln x+ 3x^2 - 4)}x.$$
A: If you wish to use logarithmic differentiation, you can proceed as follows:
\begin{gather*}
  \ln y = \frac{x^2}{3x+2} \\
  \frac{1}{y}y' = \frac{(3x+2)(2x) - x^2(3)}{(3x+2)^2} \\
  \frac{1}{y}y' = \frac{3x^2 + 4x}{(3x+2)^2} \\
  y' = y\frac{3x^2+4x}{(3x+2)^2} = e^{x^2/(3x+2)}\frac{3x^2+4x}{(3x+2)^2}.
\end{gather*}
However, it is just as easy, if not easier, to simply use the chain rule:
\begin{equation*}
  \frac{d}{dx}\left(e^{x^2/(3x+2)}\right) 
     = e^{x^2/(3x+2)}\frac{d}{dx}\left(\frac{x^2}{3x+2}\right).
\end{equation*}
A: You don't have to take any logarithms, just use the chain rule:
$$\frac{d}{dx}e^x=e^x$$
$$\frac{d}{dx}\frac{x^2}{3x+2}=\frac{3x^2+4x}{(3x+2)^2}$$
Hence:
$$\frac{d}{dx}e^{\frac{x^2}{3x+2}}=\frac{3x^2+4x}{(3x+2)^2}e^{\frac{x^2}{3x+2}}$$
If you insist:
$$\ln(y(x))=\frac{x^2}{3x+2}$$
Differentiating both sides gives:
$$\frac{y'(x)}{y(x)}=\frac{3x^2+4x}{(3x+2)^2}$$
Hence
$$y'(x)=\frac{3x^2+4x}{(3x+2)^2}y(x)=\frac{3x^2+4x}{(3x+2)^2}e^{\frac{x^2}{3x+2}}$$
A: The point is using the chain rule together. I'll do the second one for you, and you try the first one. $y = -10x^{3x^2}-4 \implies y+4 = -10x^{3x^2} \implies \ln(y+4) = 3x^2\ln(-10x).$
Now we notice that $(y+4)' = y'$, and apply $\frac{\rm d}{{\rm d}x}$ on both sides: $$\frac{\rm d}{{\rm d}x} \left(\ln(y+4)\right)= \frac{\rm d}{{\rm d}x}\left(3x^2\ln(-10x)\right) \implies \frac{1}{y+4} \frac{{\rm d}y}{{\rm d}x} = 6x\ln(-10x) + 3x^2 \frac{(-10)}{(-10x)},$$ where we used the chain rule and my previous remark in the LHS, and the usual product rule in the RHS. Proceeding: $$\frac{{\rm d}y}{{\rm d}x} = (y+4)\left(6x\ln(-10x)+3x\right) \implies \frac{{\rm d}y}{{\rm d}x} =(-10x^{3x^2})\left(6x\ln(-10x)+3x\right).$$
A: 
Do the second one in a similar fashion.
