Calculus! #37: Early Transcendentals 3.5A

Section 3.5: Implicit Differentiation

WOOP!

Implicit differentation is actually one of the concepts that got me into the idea of textblogging. It’s something that completely baffled me the first time around because I didn’t really understand what a derivative was. Later, when I sat down and made myself understand, the simplicity and utility of implicit differentation became clear.

So, first up, what does “implicit” mean here?

It’s actually a very simple idea. Heretofore we’ve only dealt with functions that look like this: f(x) = some operations being done to x. For example, y = sin(x). That is, we’ve always had functions where y is on the left, and everything to do with x is on the right. That is “explicit” form. “Implicit” form is anything else.

(Note: In this section, for ease of explanation, I’ll switch to using y instead of f(x). They mean the same thing.)

For example of implicit form, what if you have an equation like $x^3 + y^3 = 6xy$ ?. There’s probably some way you could get that to be in y = something(x) format, but it’d be ugly and cumbersome.

But say you could still take the derivative (that is, find how y changes as x changes) without changing the original equation? This is called “implicit differentiation.”

So, let’s go back to our ugly equation: $x^3 + y^3 = 6xy$. And let’s ask each chunk of each side of that equation “how do you change as x changes?” Remember, the way we do this is we apply a little d/dx to each member. Algebra tells us we can always do the same thing to both sides of the equation, and Calculus tells us we can distribute the d/dx across the two added expressionso on the left.

$\dfrac{d}{dx}x^3 + \dfrac{d}{dx} y^3 = \dfrac{d}{dx} 6xy$

The first one is simple. It’s just $3x^2$

The middle one is trickier. We’re going to use the chain rule. So, the outer function will be $y^3$, and the inner will be y.

The derivative of the first is just $3y^2$. That may seem a little like cheating, but remember, y=f(x). So, y is still a function of x, and so it still make sense that you can take the derivative with respect to x.

The derivative of the second part (and here’s where it gets cute) is just d/dx of y, also known as dy/dx, also known as y’, also known as the derivative of y with respect to x, also known as the thing you want to know when you’re solving for the derivative!

On the right side, we simply apply the product rule and end up with 6xy’ + 6y.

Put that all together, and you’ve got $3x^2 + y'3y^2 = 6xy' + 6y$.

You can simplify that to a form where y’ is all on one side, and you’re all good to go. Pretty cool, right?

You may also notice a rule that’ll save you a little time: For a power function that is base-f(x), just take the derivative like it’s x and multiply by f’(x).

This is another section where it’s important to work some problems. That said, if you understand what it means do run the operation denoted by d/dx, it should all be pretty straightforward. Just use the trick you learned in this section in concert with the chain and product rules. Niiiiice and easy.

I want to add that, now that we’ve got a few tools, you may notice a shift in the “feel” of sections for a bit. We’re now done with the bare foundation and have started to build the tower. In order to do this section, you’ve got to be able to rely on a functional understanding of what came before. This will continue for a bit – for the next 1.5 chapters, we’re building up the toolkit.

Next stop: Derivatives of Inverse Trig functions (DON’T BE AFRAID!)

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2 Responses to Calculus! #37: Early Transcendentals 3.5A

1. Jacob says:

I’m afraid.

2. Lccb says:

Nice explanation, I never understood implicit differentiation completely but this makes it a little easier to start doing exercises until grasping the concept.