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Derivatives of inverse functions: from equation | AP Calculus AB | Khan Academy

3m read
·Nov 11, 2024

Let ( F(x) ) be equal to ( 12x^3 + 3x - 4 ). Let ( H ) be the inverse of ( F ). Notice that ( F(-2) ) is equal to (-14) and then they're asking us what is ( H'(-14) ).

If you're not familiar with how functions and their derivatives relate to their inverses, well, this will seem like a very hard thing to do. If you attempt to take the inverse of ( F ) to figure out what ( H ) is, it will be tough to find, to take, to figure out the inverse of a third-degree polynomial defined function like this.

So, the key property to realize is that if ( F ) and ( H ) are inverses, then ( H'(x) ) is going to be equal to ( \frac{1}{F'(H(x))} ). You could now use this in order to figure out what ( H'(-14) ) is.

Now, I know what some of you are thinking, because it's exactly what I would be thinking if someone just sprung this on me: where does this come from? I would tell you this comes straight out of the chain rule.

We know that if we have a function and its inverse, that ( F(H(x)) ) is equal to ( x ). This literally comes out of them being each other's inverses. We could have also said ( H(F(x)) ) will also be equal to ( x ). Remember, ( F ) is going to map, or ( H ) is going to map from some ( x ) to ( H(x) ), and then ( F ) is going to map back to that original ( x ). That’s what inverses do. So, they are inverses; this is by definition.

But then if you took the derivative of both sides of this, what would you get? Let me do that. If you take the derivative of both sides, ( \frac{d}{dx} ) on the left-hand side and ( \frac{d}{dx} ) on the right-hand side, and I think you see where this is going.

You're essentially going to get a version of that. The left-hand side, using the chain rule, you're going to get ( F'(H(x)) \cdot H'(x) ) straight out of the chain rule is equal to the derivative of ( x ), which is just going to be equal to one. Then you divide both sides by ( F'(H(x)) ) and you get our original property there.

So now, with that out of the way, let's just actually apply this. We want to evaluate ( H'(-14) ).

Now, have they given us ( H(-14) )? Well, they didn't give it to us explicitly, but we have to remember that ( F ) and ( H ) are inverses of each other. So if ( F(-2) ) is ( -14 ), well, ( H ) is going to go from the other way around. If you input ( -14 ) into ( H ), you're going to get ( -2 ). So ( H(-14) ) is going to be equal to ( -2 ).

Once again, they are inverses of each other. So ( H(-14) ) is equal to ( -2 ). That's what the inverse function will do. If ( F ) goes from ( -2 ) to ( -14 ), ( H ) is going to go from ( -14 ) back to ( -2 ).

Now we want to evaluate ( F'(-2) ). Let’s figure out what ( F'(-2) ) is.

So, ( F'(x) ) is equal to ( 36x^2 + 3 ). We’re just going to leverage the power rule. So ( 3 \times 12 ) is ( 36 ) multiplied by ( x^{3-1} ), which is just ( x^2 ), plus the derivative of ( 3x ) with respect to ( x ). Well, that's just going to be ( 3 ).

The derivative of a constant is just going to be zero, so that’s ( F'(x) ). So ( F'(-2) ) is going to be ( 36(-2)^2 + 3 ).

Calculating that gives us ( 36 \times 4 + 3 ) which is ( 144 + 3 ), so that's equal to ( 147 ).

So, this denominator right here is going to be equal to ( 147 ), and this whole thing is equal to ( \frac{1}{147} ).

This was a, you know, this isn’t something you’re going to see every day. This isn’t a typical problem in your calculus class, but it's interesting.

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