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The question is:

Let $G(x,y)=x^2y^4-3x^4y$.

(i) Find the first partial derivatives $G_x$ and $G_y$.

(ii) Using (i) above, find $\frac{dy}{dx}$.

(iii) If $G(x,y)=0$, confirm your answer in part (ii) above, finding $\frac{dy}{dx}$ using implicit differentiation.

Zilliput
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Terina Gerber Swanepoel
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1 Answers1

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The chain rule says: $$ dG = \frac{\partial G}{\partial x} dx + \frac{\partial G}{\partial y}dy. $$ If the point $(x,y)$ moves along a level set of $G$, then we have $dG=0$. Hence $$ \frac{dy}{dx} = \frac{-\partial G/\partial x}{\phantom{-}\partial G/\partial y}. $$ $$ = -\frac{2xy^4 - 12x^3 y}{x^2 4y^3 -3x^4} $$ and then we can cancel an $x$.

Now let's try implicit differentiation: $$ x^2y^4 - 3x^4y = 0. $$ $$ 2x y^4 + x^2 4y^3 \frac{dy}{dx} - 12x^3y - 3x^4\frac{dy}{dx} =0. $$ Push the two terms not involving the derivative to the other side; then pull out the common factor, which is the derivative; then divide both sides by the other factor. We get $$ \frac{dy}{dx} =\frac{12x^3y - 2xy^4}{x^24y^3 - 3x^4} $$ and it's the same thing.

Michael Hardy
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