Take the derivative of another variable

Question

How can you take the derivative

$$\frac{d}{dx}(y^2)$$

I don't understand how the chain rule applies here. Someone told me that the chain rule applies here because $y$ can be expressed in some type of $x$ expression, but I still don't get it.

So, I do understand how to apply the chain rule, I just don't understand why it applies in this situation.

I checked this video.

Answer 1

Notice that by the Chain Rule you have $g(f(x)) = g'(f(x))f'(x)$. Now assuming $y(x)$ is a function of $x$ we have that

$$\frac{d}{dx} (y^2) = 2y\ \frac{dy}{dx} $$

Take $g(y) = y^2$ and $y = f(x)$.

Answer 2

$$ u(y) = y^2\\ y = f(x). $$ Thus we have by chain rule $$ \frac{du}{dx} = \frac{dy}{dx}\frac{du}{dy} = \frac{dy}{dx}\dfrac{d}{dy}y^2 = 2y'y $$ So we have that $$ \frac{du}{dx} =\dfrac{d}{dx}y^2 =2y'y $$

Answer 3

The chain rule always applies. You derive a function of $y$ (namely $y^2$) with respect to $x$, which is the same as deriving $y^2$ with respect to $y$ and multiplying by the derivative of $y$ with respect to $x$.

$$\frac{dy^2}{dx}=\frac{dy^2}{dy}\cdot\frac{dy}{dx}=2y\frac{dy}{dx}.$$

Whether there is a dependency between $y$ and $x$ doesn't matter, the rule remains true.

If there is no dependency,

$$\frac{dy^2}{dx}=2y\frac{dy}{dx}=2y\cdot0.$$

Answer 4

I see this as differentiating with respect to $y$ and making up for it with a scale factor $\frac{dy}{dx}$ $$\frac{d}{dx}\{\}=\frac{dy}{dx}\frac{d}{dy}\{\}$$

Hence:$$\frac{d}{dx}\{y^2\}=\frac{dy}{dx}\times 2y$$