# Explain why the chain rule is needed to find derivatives if $y$ is a function of $x$?

I know that if $y$ is a function of $x$, or $y=f(x)$, you need to use the chain rule to find it's derivative.

Let's say I want to find the derivative of $y^2$ and $y$ is a function of $x$. Therefore, I would need to use the chain rule. What are the inside and outside functions of $y^2$ using the chain rule (Both terms would be functions of $x$)?

I also want to know why the derivative of $y^2$ is the same as the derivative of $2y$, by using the chain rule.

First, "I also want to know why the derivative of $y^2$ is the same as the derivative of $2y$, by using the chain rule." --> They aren't? I'm not sure what you mean here but $2y$ is the derivative of $y^2$ with respect to $y$. They're clearly not the same function and unless $y$ is trivial, you won't get the same thing when you take the derivatives of $y^2$ and $2y$.

Second, it's very important to specify what you're taking a derivative with respect to when you have multiple variables in play.

For example, let's say $y = x^2 + 5x$ and I want to take the derivative of $y^2$. This is an ambiguous statement. If you mean the derivative of $y^2$ with respect to $y$, then you would get $2y$. If you mean the derivative of $y^2$ with respect to $x$, then this is where the chain rule comes into play. The chain rule tells us that derivatives kind of work like fractions:

$$\frac{d}{dx}(y^2) = \frac{dy}{dx} \frac{d}{dy}(y^2) = \frac{dy}{dx} (2y) = (2x+5)(2y) = 2(2x+5)y = 2(2x+5)(x^2+5x)$$

The "inside function" would be $x^2+5x$ -- the derivative of this with respect to $x$ is the $\frac{dy}{dx}$ part. The "outside function" would be $y^2$ -- the derivative of this with respect to $y$ is the $\frac{d}{dy}(y^2)$.

It depends on what you are really dealing with. If y is a function of x and z is a function of y, as in $y=f(x)$ and $z=y^2$ then $$\frac{dz}{dy}=2y,\text{ but }\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}=2yy'=2f(x)f'(x).$$

Not true that both inside and outside functions are functions of $x$. $y^2$ is a function of $y$. It's helpful to take a concrete example for $y=f(x)$. Take $y=\sin x$ for example. Then $y^2=(f(x))^2=\sin^2x$. Now work through the chain rule.

This only involves an application of the chain rule once. Let $y=f(x)$, then these things are the same: $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}f}{\mathrm{d}x}=f'(x)$$

Now, let $g(x)=x^2$. Recall that $g'(x)=2x$. What you want to find is the derivative of $g(f(x))$ with respect to $x$. This means we can use the chain rule:

$$\frac{\mathrm{d}}{\mathrm{d}x} g(f(x)) = g'(f(x))f'(x) = \boxed{2f(x)f'(x)}$$

An example: suppose $y=f(x)=x^3 + 1$. What is the derivative of $y^2$ with respect to $x$?

$$\frac{\mathrm{d}}{\mathrm{d}x} y^2 = 2\left(x^3+1\right)\left(3x^2\right) = 6x^5 + 6x^2$$

We can also find this derivative of $y^2$ by expanding: $$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^3+1\right)^2 = \frac{\mathrm{d}}{\mathrm{d}x} \left(x^6 + 2x^3 + 1\right) = 6x^5 + 6x^2$$

This last method doesn't require the chain rule at all, but requires that it is easy to both apply $g$ to $f$ and then take the resulting derivative directly.

• Makes sense, but which function is the inner and which function is the outer? Mar 9, 2015 at 22:46
• $f$ is the inner function, and $g$ is the outer function. Mar 9, 2015 at 22:48