Implicit differentiation at a given point

Question

I was given this problem on my online homework, but I'm getting stuck when trying to get the $dy/dx$ term on one side.

The question is: "For the equation given below, evaluate $dy/dx$ at the point $(1, -5/11)$.

$$\frac{y}{x+2y}=x^4-6$$

$dy/dx$ at $(1, -5/11)$=?"

I did some work on it, but I don't know where to go from where I left off. Also, I'm using $y'$ instead of $dy/dx$ for simplicity. Here is what I have so far:

$$y/(x+2y)=x^4-6$$

Using the quotient rule on the left side and power rule on right:

$$\begin{eqnarray} \frac{((y')(x+2y)-(y)(1+2y')}{(x+2y)^2}=4x^3\\ \frac{(xy'+2yy'-(y+2yy')}{(x+2y)^2}=4x^3\\ \frac{xy'-y}{(x+2y)^2}=4x^3\\ \end{eqnarray}$$ Then I tried splitting the left side up so I could move some of it to the right:

$$\frac{xy'}{(x+2y)^2}-\frac{y}{(x+2y)^2}=4x^3$$

Moved some to the right:

$$\frac{xy'}{(x+2y)^2}=4x^3-\frac{y}{(x+2y)^2}$$

I'm stuck at this point, and I'm not even sure what I did was correct so any feedback would help. Am I on the right track?

Answer 1

Hint: $$y/(x+2y)=x^4-6\\ \implies \dfrac{(x+2y)y^{'}-y(1+2y^{'})}{(x+2y)^{2}}=4x^3\\ \implies \dfrac{4x^3(x+2y)^{2}+y}{x}=y^{'}$$

Answer 2

Only the title of your post tells about implicit differentiation. If this is not a requirement, you can extract $y$ from the equation and get $$y=-\frac{x \left(x^4-6\right)}{2 x^4-13}$$ from which $$y'=\frac{-2 x^8+29 x^4-78}{\left(13-2 x^4\right)^2}$$ You can check that for $x=1$, $y=-\frac{5} {11}$ and, at this point $y'=-\frac{51} {121}$