Dtjytyk

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

$$frac{d}{dx} (11x^5)=11frac{d}{dy}(5x^4) = 55x^4$$

$$frac{d}{dx} (x^4y) = [4x^3 cdot y] + [1 cdot x^4] = 4yx^3+x^4$$

$$frac{d}{dx}(xy^5) = [1 cdot y^5] + [x cdot 5y^4] = y^5 + 5xy^4$$

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:
$$(x^4y)'=x^4(y)'+(x^4)'y=x^4frac{dy}{dx}+4x^3y$$
$$(xy^5)'=x(y^5)'+(x)'y^5=5xy^4frac{dy}{dx}+y^5$$
Thus
$$55x^4+x^4frac{dy}{dx}+4x^3y+5xy^4frac{dy}{dx}+y^5=0$$
$$(x^4+5xy^4)frac{dy}{dx}=-(55x^4+4x^3y+y^5)$$
$$frac{dy}{dx}=-frac{55x^4+4x^3y+y^5}{x^4+5xy^4}$$

differentiating each term

$$frac{d}{dx} (11x^5)=11(5x^4) = 55x^4$$

$$frac{d}{dx} (x^4y) = [4x^3 cdot y] + [1 cdot x^4color{red}{frac{dy}{dx}}] $$

$$frac{d}{dx}(xy^5) = [1 cdot y^5] + [x cdot 5y^4color{red}{frac{dy}{dx}}] $$

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:

$55x^4+yfrac{d}{dx}x^4+x^4frac{dy}{dx}+y^5frac{d}{dx}x+xcdot4y^4frac{dy}{dx}=0$

$55x^4+ycdot3x^3+x^4frac{dy}{dx}+y^5+4xy^4frac{dy}{dx}=0$

$55x^4+3x^3y+y^5+frac{dy}{dx}(x^4+4xy^4)=0$

$frac{dy}{dx}(x^4+4xy^4)=-55x^4-3x^3y-y^5$

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

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 $frac{d}{dx}(y^5)= 5y^4frac{dy}{dx}$ exactly because $y = y (x)$. Then

$$frac{d}{dx}(11x^5 + x^4y+xy^5) = frac{d}{dx}(18)$$

$$frac{d}{dx}(11x^5)+frac{d}{dx}(x^4y) + frac{d}{dx}(xy^5)=0$$

$$(55x^4)+ left( 4x^3 y + x^4 frac{dy}{dx} right) +left( y^5+5xy^4frac{dy}{dx}right) = 0 ,,,,,,,,,,(*)$$

$$left( x^4 frac{dy}{dx} right) +left(5xy^4frac{dy}{dx}right) = -55x^4-4x^3 y- y^5$$

$$left( x^4 +5xy^4right)frac{dy}{dx} = -55x^4-4x^3 y- y^5$$

$$frac{dy}{dx} = dfrac{-55x^4-4x^3 y- y^5}{left( x^4 +5xy^4right)}$$

I think that it is easy to remember the implicit function theorem.
$$color{red}{F(x,y)=0 implies frac{dy}{dx}=-frac{frac {partial F(x,y)} {partial x}} {frac {partial F(x,y)} {partial y}}}$$

$$F(x,y)=11x^5 + x^4y + xy^5-18=0$$
$$frac {partial F(x,y)} {partial x}=55x^4+4x^3y+y^5$$
$$frac {partial F(x,y)} {partial y}=x^4+5xy^4$$
$$frac{dy}{dx}=-frac{55x^4+4x^3y+y^5 } {x^4+5xy^4 }$$