# Implicit derivative: why do we keep the $\frac{dy}{dx}$?

I just started learning calculus and I'm studying implicit derivatives and I have a question regarding the differenciation of the y variable. I'll use an example:

Applying implicit derivative to $$5y^2 = x^2$$

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

$$10y \frac{dy}{dx} = 2x$$

Why do we keep the $$\frac{dy}{dx}$$ after differentiating $$5y^2$$?

The book I'm following does not explain the reason for this.

• it is the chain rule – Dr. Sonnhard Graubner Apr 22 at 14:43
• You can see why this is necessary if you notice the following. In reality, there is also a $\frac{dx}{dx}$ on the right-hand side as well; however, since the rate of change of a variable with respect to itself is unity, $\frac{dx}{dx}x^2 = x^2$. – Victoria M Apr 22 at 14:44

In this problem on implicit differentiation you are thinking about how the value of $$y$$ depends on the value of $$x$$, without explicitly finding an equation for the function $$f$$ so you could write $$y = f(x)$$. You can rewrite the left side of the equation in terms of the unknown function $$f$$ this way: $$5(f(x))^2.$$ To differentiate that you need the chain rule $$\frac{d}{dx}5(f(x))^2 = 5\times 2 f(x)f'(x) = 10 y \frac{dy}{dx}.$$

If you are comfortable thinking like a physicist or have developed your mathematical intuition about small changes you can reason with differentials:

$$dQ$$ is the infinitesimal change in quantity $$Q$$ caused by some infinitesimal change elsewhere.

Now suppose quantities $$x$$ and $$y$$ are related by the equation $$5y^2 = x^2 .$$ If $$x$$ changes by $$dx$$ then $$y$$ changes by $$dy$$ and the chain rule says, essentially $$d(5y^2) = 5 \times 2y \ dy = 2x \ dx .$$ You can solve this equation for the ratio $$dy/dx$$.

Mathematicians think this way for themselves all the time. Do not do it in your elementary calculus class unless your instructor says its OK.

• Your explanation was very clear. Thank you. – gmadalosso Apr 22 at 18:05

What would you expect the derivative of $$5y^2$$ with respect to $$x$$ to be?

In your implicit differentiation you are assuming that $$y$$ is a function of $$x$$. To differentiate $$5y^2$$ with respect to $$x$$, you need to apply the Chain Rule.

Think of $$y$$ as representing a function of $$x$$, as in $$y(x)$$. Therefore, the equation is essentially$$5(y(x))^2=x^2$$

Therefore, differentiating with respect to $$x$$, we use the chain rule on $$y(x)$$10y(x)y'(x)=2x

$$\frac{dy}{dx}$$ is typically what you are looking for; it is the derivative. You can solve the equation (which you found by implicitly differentiating) for $$\frac{dy}{dx}$$ (in your example you can solve the equation for the derivative by dividing both sides by $$10y$$).

This is no different than finding the derivative of an explicit equation except that there's some algebra left to do at the end. Even when deriving an explicit equation you're "keeping" $$\frac{dy}{dx}$$ there.

When you derive $$y=x^2$$, you get $$\frac{dy}{dx} = 2x$$. You're always keeping the $$\frac{dy}{dx}$$ there when you derive with respect to $$y$$ $$:)$$

$$\frac{d}{dy}y^2=2y$$

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