# Quotient Rule Proof – Calculus

by Batool Akmal

My Notes
• Required.
Learning Material 2
• PDF
DLM Quotient Rule, Chain Rule and Product Rule Calculus Akmal.pdf
• PDF
Report mistake
Transcript

00:01 So after proving chain rule and product rule, this should be fairly easier so we're now going to look at the proof for the quotient rule.

00:09 In order to do this we are going to use the product rule because it's a bit more simpler and we're just going to apply the product rule and hopefully derive the same result that we have for the quotient rule.

00:20 So let?s start to look at our proof for the quotient rule.

00:24 So we're trying to prove the derivative of f of x over g of x and we're trying to prove that the derivative does actually give us the quotient rule.

00:38 So this is what we are trying to do. We can rewrite this and bring the g up.

00:44 So remember the g of x has a power of one, so we can bring that up and we can rewrite this as d/dx, f of x, multiplied by g of x to the minus one.

00:55 So all I've done is I have brought g of x up and changed its power because it was plus one at the bottom, it now becomes a minus one.

01:03 Okay so we're trying to differentiate a function, so we're trying to differentiate this but also observe this closely.

01:10 This is a product of two functions so in order to differentiate it we can use the product rule.

01:17 Remember what the product rule stated so the product rule said, if we just write by here, dy/dx you can do this in any order that you want but you can do it as uv dash plus vu dash so leave one as it is, differentiate the other and differenti-- leave the other second one as it is and differentiate the first.

01:34 So if we start with just differentiating f of x ?cause it?s easier, you can leave g, x of minus 1 as it is.

01:42 We now treat this as a chain rule function so we're now going to leave f of x as it is and we're now going to differentiate g. In order to do that you bring the power down first, so like you generally do. You decrease the power by 1 and also remember that we're going to differentiate the inside of the function so you multiply it with the differential of g dashed of x.

02:09 So that?s the chain rule applied here to this function there.

02:14 If we tidy this up a little bit and we write this we can write this as f dashed of x and you can take this back down, so you can take this term back down to make it positive, divided by g of x.

02:25 We now have a minus that multiplies here so instead of a plus we're going to get a minus there.

02:31 We have f of x, g dash of x, and you can take g x squared at the bottom so it?s this term here when it goes down by the rules of indices minus 2 will now become positive.

02:47 We're almost there. We can add these two fractions by taking a common denominator.

02:53 So we can take a common denominator of g x squared.

02:58 So this is just basic algebra now. In order to add the functions you want to increase this function here with the g of x because you want g of x squared so you can have f dash of x, g of x, the minus in the inside stays.

03:13 This term doesn?t need to change because it?s already on g of x squared, so we just get f of x g dash of x, and if you look at this closely, you should feel fairly pleased with yourselves because we have derived the quotient rule, so we have here vdudx, so we have vdudx, minus udvdx, all over v squared, if you call your f of x, u and your g of x, v.

03:43 So well done at deriving not only the quotient rule but all the other proofs so we've proved chain rule and the product rule.

03:53 Again these are incredible algebraic skills and mathematical skills and derivation skills which will surely help you develop further especially in the mathematical side of your medical degrees.

The lecture Quotient Rule Proof – Calculus by Batool Akmal is from the course Quotient Rule, Chain Rule and Product Rule.

### Included Quiz Questions

1. ( h'(x)g(x) - 2g'(x)h(x) )/g(x)³
2. ( h'(x)g(x) - 2g'(x)h(x) )/g(x)⁴
3. ( h'(x)g(x) - 2g'(x)g(x)h(x) )/g(x)⁴
4. ( h'(x)g(x) + 2g'(x)h(x) )/g(x)³
5. ( h'(x)g(x)² - 2g'(x)h(x) )/g(x)⁴
1. The product rule and the chain rule
2. The product rule
3. The chain rule
4. Differentiation by substitution and the product rule
5. The chain rule and the power rule
1. dy/dx = (12x² + 40x + 3)/(1-4x²)²
2. dy/dx = (24x² + 40x + 3)/(1-4x²)²
3. dy/dx = (-36x² - 40x + 3)/(1-4x²)²
4. dy/dx = (12x² + 40x + 3)/(1-4x²)
5. dy/dx = (-24x² - 40x - 3)/(1-4x²)²

### Customer reviews

(1)
5,0 of 5 stars
 5 Stars 5 4 Stars 0 3 Stars 0 2 Stars 0 1  Star 0