Let F And G Be Differentiable Functions Such That: Complete Guide

You’re Looking at Two Functions. Now What?

So you’re staring at a problem. Because of that, they’re differentiable. Plus, just a statement. Here's the thing — cool. It says: let f and g be differentiable functions. So just… two functions. That’s it. No graph, no numbers, no specific formulas. What do you do with that?

If you’ve ever felt a wave of abstract dread reading this, you’re not alone. Also, it’s the starting line for half the calculus problems you’ll ever see, and it can feel like being handed a toolbox with no instructions. But here’s the thing: that simple phrase is actually a key. Plus, it unlocks a whole set of powerful tools—rules that let you handle combinations of these functions, no matter what they are. It’s less about these specific f and g, and more about any f and g that play by the differentiable rules.

This isn’t just academic trivia. Understanding what “differentiable” guarantees—and what you can build from it—is one of the most practical superpowers in applied math, physics, economics, and engineering. The growth of a population (f) competing for a limited resource (g). The speed of a car (one function) affected by wind resistance (another function). And this is how you model real, messy change. Let’s stop staring at the toolbox and start opening it.

What “Let f and g Be Differentiable Functions” Actually Means

Okay, dictionary definition time—just kidding. No sharp corners, no sudden jumps, no vertical cliffs. Plus, it means both f(x) and g(x) are smooth. Day to day, if you could draw them without ever lifting your pencil from the paper (and without any cusps), they’re differentiable. We’re not doing that. In plain English? More technically, at every point in their domain, they have a well-defined, finite slope—a derivative, f’(x) and g’(x).

The “let” part is the magic. We don’t need to know their exact formulas because the rules we’re about to apply will work for any formulas that meet this smoothness condition.” It’s a universal pass. It’s a mathematician’s way of saying: “We’re going to work with two general, smooth functions. You’re not solving for one answer; you’re proving a relationship that holds for a whole category of functions.

Why This Simple Phrase Is the Foundation of Half of Calculus

Why should you care about this generality? Because life doesn’t give you neat, single-function problems. Reality is combinations.

Think about it:

• Your total profit might be revenue (one function) minus costs (another function). Day to day, * The volume of a heated gas might depend on pressure (one function) and temperature (another function) interacting. * The position of a robot arm might be a function of two different motor angles.

In each case, you’re not just dealing with f or g alone. You’re dealing with f + g, f * g, f(g(x)), or f(x)/g(x). The statement “let f and g be differentiable” is your green light. Think about it: it guarantees that these combinations will also be differentiable (with a few important caveats). And if they’re differentiable, you can find their derivatives using a standard set of rules. You can analyze their rates of change. You can optimize them. You can model their behavior. Without that guarantee of smoothness for f and g, the whole elegant system falls apart.

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How It Works: The Rulebook for Combining Smooth Functions

This is the meat. Even so, given that f and g are differentiable, here’s what you can do. This is the rulebook your professor assumes you know.

The Sum and Difference Rule: (f ± g)’ = f’ ± g’

This one’s beautifully intuitive. The derivative of a sum is the sum of the derivatives. If you’re tracking two smooth changing quantities, the rate of change of their total is just the sum of their individual rates of change.

• Example: Your total distance traveled is forward distance (f) minus backward distance (g). Your net speed is simply forward speed (f’) minus backward speed (g’).

The Product Rule: (f * g)’ = f’ * g + f * g’

Here’s where intuition can trip you up. It’s not just f’ times g’. You have to account for both parts changing. Think of it as: “The change in the product comes from (f changing while g stays put) plus (g changing while f stays put).”

• The mnemonic “lo d-hi minus hi d-lo over lo-lo” for quotients is famous, but for product, I just think: first times derivative of second, plus second times derivative of first.
• Why it matters: If revenue is price (f) times quantity sold (g), your marginal revenue isn’t just (change in price) * quantity. It’s (change in price)current quantity + current price(change in quantity). Both effects matter.

The Quotient Rule: (f / g)’ = (f’ * g – f * g’) / g²

This one’s clunkier, but it’s just the product rule in disguise (since f/g = f * (1/g)). The key thing the “differentiable” assumption gives you here is that g(x) ≠ 0 in the domain you care about, and also that 1/g is differentiable (which requires g ≠ 0). The formula is what it is. Drill it.

• Big Warning: The denominator is g squared, not just g. And the numerator is a specific subtraction: (f’ * g) – (f * g’). Order matters. Flip it and you’re wrong.

The Chain Rule: The Big One

This is the powerhouse. It’s for when you have a function inside another function: h(x) = f(g(x)). The chain rule says: h’(x) = f’(g(x)) * g’(x).

• In words: Differentiate the outer function, evaluated at the inner function, then multiply by the derivative of the inner function.
• It’s why “differentiable” is so crucial. For f(g(x)) to be differentiable, we need g to be differentiable (so its output changes smoothly) and we need f to be differentiable at the points g(x) (so it can handle the smooth input from g).
• Example: You’re heating a metal rod. The temperature T depends on time t: T(t). But