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How do you differentiate this equation with respect to $x$? $$x^2=xy^3+2$$

$$\frac{d}{dx}(x^2)=\frac{d}{dx}(xy^3)+\frac{d}{dx}(2)$$

$$2x=x\frac{d}{dx}(y^3)+y^3\frac{d}{dx}(x)+0$$

$$2x=x\frac{d}{dx}(y^3)+y^3$$

Here is the problem I am facing with, $$\frac{d}{dx}(y^3)?$$

this $\frac{d}{dx}(x^3)=3x^2$ it is understandable

2 Answers2

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Use the chain rule and get

$\displaystyle \frac{d}{dx}y^3 = 3y^2\frac{dy}{dx}$

bob.sacamento
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Recall that $$ \frac{d}{dy} \cdot \frac{dy}{dx}= \frac{d}{dx} $$ by the chain rule. Then we have $$ \frac{d}{dx}y^3 = \frac{dy}{dx} y^3 \cdot \frac{d}{dy} = 3y^2 \frac{dy}{dx}. $$ So $$ \frac{d}{dx} y^3 = 3y^2\frac{dy}{dx}.$$ If you are interested in what the value of $\frac{dy}{dx}$ is, then you can solve for it as if it was a variable. It should come out to be a function in terms of $x$ and $y$.

Jane Cooper
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