Differentiate $11x^5 + x^4y + xy^5=18$ I am not sure how to differentiate $11x^5 + x^4y + xy^5=18$. I have a little bit of experience with implicit differentiation, but I'm not sure how to handle terms where both variables are multiplied together.
I have tried
$$\frac{d}{dx}(11x^5 + x^4y+xy^5) = \frac{d}{dx}(18)$$
$$\frac{d}{dx}(11x^5)+\frac{d}{dx}(x^4y) + \frac{d}{dx}(xy^5)=0$$
differentiating each term
$$\frac{d}{dx} (11x^5)=11\frac{d}{dy}(5x^4) = 55x^4$$
$$\frac{d}{dx} (x^4y) = [4x^3 \cdot y] + [1 \cdot x^4] = 4yx^3+x^4$$
$$\frac{d}{dx}(xy^5) = [1 \cdot y^5] + [x \cdot 5y^4] = y^5 + 5xy^4$$
finding $\frac{dy}{dx}$
$$\frac{dy}{dx}=\frac{-x}{y} = \frac{-[4yx^3+x^4] + [y^5+5xy^4]}{55x^4}$$
According to the website I'm using, "WebWork", this is wrong.

For each term involving both $x$ and $y$, differentiate the $x$ and $y$ parts separately and combine the two using the product rule:
$$(x^4y)'=x^4(y)'+(x^4)'y=x^4\frac{dy}{dx}+4x^3y$$
$$(xy^5)'=x(y^5)'+(x)'y^5=5xy^4\frac{dy}{dx}+y^5$$
Thus
$$55x^4+x^4\frac{dy}{dx}+4x^3y+5xy^4\frac{dy}{dx}+y^5=0$$
$$(x^4+5xy^4)\frac{dy}{dx}=-(55x^4+4x^3y+y^5)$$
$$\frac{dy}{dx}=-\frac{55x^4+4x^3y+y^5}{x^4+5xy^4}$$
