Implicit differentiation with logarithm Find $y'$ if $y=\ln(7x^2+3y^2)$.
I'm kind of confused on this problem and could use everyone's help. Am I supposed to take the derivative first, which is $y=\ln(14x+6y)$? If so, how do I go from there?
 A: $\def\ddx{\frac{d}{dx}}$First, if you are taking the derivative you have to do it to both sides - you did it for the RHS but you didn't do anything with the LHS.  We have
$$\ddx(LHS)=\ddx(RHS)\ ,$$
so
$$y'=\ddx(RHS)\ .$$
Second problem is that you did not find the derivative of the RHS correctly.  You will need to use the chain rule (in fact you will need to use it twice).  You should know that
$$\ddx(\ln x)=\frac{1}{x}$$
as long as $x$ is positive, so if $u$ is a function of $x$ then by the chain rule
$$\ddx(\ln u)=\frac{1}{u}\frac{du}{dx}\ .$$
If you take $u=7x^2+3y^2$ then
$$\eqalign{\ddx\ln(7x^2+3y^2)
  &=\frac{1}{7x^2+3y^2}\ddx(7x^2+3y^2)\cr
  &=\frac{1}{7x^2+3y^2}\Bigl(14x+\ddx(3y^2)\Bigr)\ .\cr}$$
Now note that the derivative of $3y^2$ with respect to $y$ is $6y$, however we are not differentiating with respect to $y$ but with respect to $x$.  So we have to use the chain rule again: if $v=y^2$ then
$$\ddx(y^2)=\frac{dv}{dx}=\frac{dv}{dy}\frac{dy}{dx}=6yy'\ .$$
Putting everything together, we have
$$y'=\frac{1}{7x^2+3y^2}(14x+6yy')\ :$$
now treat this as an equation where $y'$ is the unknown, and solve to find an expression for $y'$ in terms of $x$ and $y$.
With a bit of practice you should be able to write down this last equation just by looking at the question, but in the early stages it is probably best to give full details of the working.  Good luck!
A: Differentiate both sides & use the chain rule for the log with the substitution $u=7x^2+3y^2$.
$$\frac{\mathrm{d}}{\mathrm{d}x}y=\frac{\mathrm{d}}{\mathrm{d}x}\ln(7x^2+3y^2)$$
$$=\frac{\mathrm{d}}{\mathrm{d}u}\ln(u)\cdot\frac{\mathrm{d}}{\mathrm{d}x}(7x^2+3y^2)$$
Which gives us
$$y'=\frac{\mathrm{d}}{\mathrm{d}u}\ln(u)\cdot\left(\frac{\mathrm{d}}{\mathrm{d}x}7x^2+\frac{\mathrm{d}}{\mathrm{d}x}3y^2\right)$$
This should give you an equation in terms of $x$, $y$ & $y'$ which you can solve for $y'$.
