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.
 A: differentiating each term
$$\frac{d}{dx} (11x^5)=11(5x^4) = 55x^4$$
$$\frac{d}{dx} (x^4y) = [4x^3 \cdot y] + [1 \cdot x^4\color{red}{\frac{dy}{dx}}] $$
$$\frac{d}{dx}(xy^5) = [1 \cdot y^5] + [x \cdot 5y^4\color{red}{\frac{dy}{dx}}] $$
A: 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
$$55x^4+x^4\frac{dy}{dx}+4x^3y+5xy^4\frac{dy}{dx}+y^5=0$$
$$(x^4+5xy^4)\frac{dy}{dx}=-(55x^4+4x^3y+y^5)$$
$$\frac{dy}{dx}=-\frac{55x^4+4x^3y+y^5}{x^4+5xy^4}$$
A: This is how you could do it:
$11x^5+x^4y+xy^5=18$
Use the $\frac{d}{dx}$ operator (not $\frac{dy}{dx}$)
$55x^4+\frac{d}{dx}(x^4y+xy^5)=0$
Now, use the product rule on each part in the brackets:
$55x^4+y\frac{d}{dx}x^4+x^4\frac{dy}{dx}+y^5\frac{d}{dx}x+x\cdot4y^4\frac{dy}{dx}=0$
$55x^4+y\cdot3x^3+x^4\frac{dy}{dx}+y^5+4xy^4\frac{dy}{dx}=0$
$55x^4+3x^3y+y^5+\frac{dy}{dx}(x^4+4xy^4)=0$
$\frac{dy}{dx}(x^4+4xy^4)=-55x^4-3x^3y-y^5$
So, finally,
$\frac{dy}{dx}=-\frac{55x^4+3x^3y+y^5}{x^4+4xy^4}$
You were just using the operators wrong, I think
A: you actually apply the derivative with respect to the variable $x$ in both members and develop using your knowledge of the product's derivative and chain rule. Note that to calculate implicit derivatives you must think of for example $y = y (x)$. This justifies the passage I marked with (*). For examples $\frac{d}{dx}(y^5)= 5y^4\frac{dy}{dx}$ exactly because $y = y (x)$. Then
$$\frac{d}{dx}(11x^5 + x^4y+xy^5) = \frac{d}{dx}(18)$$
$$\frac{d}{dx}(11x^5)+\frac{d}{dx}(x^4y) + \frac{d}{dx}(xy^5)=0$$
$$(55x^4)+ \left( 4x^3 y + x^4 \frac{dy}{dx} \right) +\left( y^5+5xy^4\frac{dy}{dx}\right) = 0 \,\,\,\,\,\,\,\,\,\,(*)$$ 
$$\left( x^4 \frac{dy}{dx} \right) +\left(5xy^4\frac{dy}{dx}\right) = -55x^4-4x^3 y- y^5$$
$$\left( x^4  +5xy^4\right)\frac{dy}{dx} = -55x^4-4x^3 y- y^5$$
$$\frac{dy}{dx} = \dfrac{-55x^4-4x^3 y- y^5}{\left( x^4  +5xy^4\right)}$$
A: I think that it is easy to remember the implicit function theorem. 
$$\color{red}{F(x,y)=0 \implies \frac{dy}{dx}=-\frac{\frac {\partial F(x,y)} {\partial x}} {\frac {\partial F(x,y)} {\partial y}}}$$
$$F(x,y)=11x^5 + x^4y + xy^5-18=0$$
$$\frac {\partial F(x,y)} {\partial x}=55x^4+4x^3y+y^5$$
$$\frac {\partial F(x,y)} {\partial y}=x^4+5xy^4$$
$$\frac{dy}{dx}=-\frac{55x^4+4x^3y+y^5 } {x^4+5xy^4 }$$
