What comes first? Product rule/Quotient rule with chain rule

[MathNoob94]

Hi,

So I'm stumped on the concept of chain rule. I wanted to know when do I apply the quotient/product rule when working with chain rule and in which order would I apply. Mostly, when I see a question im confused on what steps I should take first. For example y=x^2sin2x when I first saw this question it confused me.


Are there any tricks or tips that can help me distinguish or identify what rule to apply first or/and when to apply the rules.

Thank You!

 

[stapel]

I'm stumped on the concept of chain rule. I wanted to know when do I apply the quotient/product rule when working with chain rule and in which order would I apply.

Work from the outside in. (In other words, work opposite from what you'd do for evaluating composed or multiplied functions.)

For example y=x^2sin2x when I first saw this question it confused me.

This is a product:

. . . . .$\displaystyle y\, =\, f(x)\, =\, \left(x^2\right)\cdot \left(\sin\left(2x\right)\right)\,=\, g(x)\cdot h(x)$

. . . . .\(\displaystyle \mbox{for }\, g(x)\, =\, x^2\, \mbox{ and }\, h(x)\, =\, \sin(2x)\)

To differentiate (from the outside inward), you first have to deal with the product:

. . . . .$\displaystyle \dfrac{df}{dx}\, =\, \left(\dfrac{dg}{dx}\right)\cdot \left(h(x)\right)\, +\, \left(g(x)\right)\cdot \left(\dfrac{dh}{dx}\right)$

To differentiate g(x), you apply the Power Rule. To differentiate h(x), you first note that this is a composed function; that is, you have one function (being polynomial multiplication) inside another function (being the sine). In other words:

. . . . .\(\displaystyle h(x)\, =\, s\left(t((x)\right)\, \mbox{ for }\, s(x)\, =\, \sin(x)\, \mbox{ and }\, t(x)\, =\, 2x\)

So here you would apply the Chain Rule.

 

[ksdhart]

When faced with a complicated, multi-step problem like that, I like to start by thinking about the steps involved, and any rules/formulas I'd need. In this case, the problem says to find $\displaystyle \frac{d}{dx}\left(x^2\cdot sin\left(2x\right)\right)$. You've identified that you'll need the chain rule and the product rule. As far as what order to do them in, you'll want to start with the chain rule, and I'll show you why:

Consider starting with the product rule. Our formula is $\displaystyle \frac{d}{dx}\left(f\cdot g\right)=\frac{d}{dx}\left(f\right)\cdot g+f\cdot \frac{d}{dx}\left(g\right)$

We have f = x^2 and g=sin(2x). So then the derivative of f is 2x, but what is the derivative of g? You can see that in order to use the product rule formula, you need to already know the derivative of g, which involves using the chain rule.

Anyway, that's just my two cents on the matter. Other users here may disagree or have further input, but I hope I was of some help to you.

 

[MathNoob94]

When faced with a complicated, multi-step problem like that, I like to start by thinking about the steps involved, and any rules/formulas I'd need. In this case, the problem says to find $\displaystyle \frac{d}{dx}\left(x^2\cdot sin\left(2x\right)\right)$. You've identified that you'll need the chain rule and the product rule. As far as what order to do them in, you'll want to start with the chain rule, and I'll show you why:

Consider starting with the product rule. Our formula is $\displaystyle \frac{d}{dx}\left(f\cdot g\right)=\frac{d}{dx}\left(f\right)\cdot g+f\cdot \frac{d}{dx}\left(g\right)$

We have f = x^2 and g=sin(2x). So then the derivative of f is 2x, but what is the derivative of g? You can see that in order to use the product rule formula, you need to already know the derivative of g, which involves using the chain rule.

Anyway, that's just my two cents on the matter. Other users here may disagree or have further input, but I hope I was of some help to you.

Thank you to everyone who replied it helped me alot. One more question although. If you see brackets in the question is that an automatic sign that chain rule is required.?

 

[Deleted member 4993]

Thank you to everyone who replied it helped me alot. One more question although. If you see brackets in the question is that an automatic sign that chain rule is required.?

No.... $\displaystyle \dfrac{d}{dx}\left [f(x) \ + \ g(x) \right ] \ = \ f'(x) \ + \ g'(x)$ does not require chain rule or quotient rule.