Chain, Product, Quotient

[kmce]

I am trying to figure out what one of the Chain, Product, Quotient equations to use with what types of questions.

Is there a standard rule that will help me figure out what one i have to use.

The types of questions I am getting are

y=x³

f(t)= 3e^sin(2t)

y=4cos(x³)

f(x)= e^x/ sinx

I need to be able to know if its the Chain, Product, Quotient to use with these, and i cannot find a simple way in which I am able to tell. I have the answers, so not looking for the answers for them, Just a way to help me realise what one to use

 

[MarkFL]

1.) \(\displaystyle y=x^3\)

The definition of \(\displaystyle y\) here has only a power of \(\displaystyle x\), so you only need to apply the power rule:

\(\displaystyle \displaystyle \frac{d}{dx}\left(x^n\right)=nx^{n-1}\)

\(\displaystyle \displaystyle \frac{dy}{dx}=3x^{3-1}=3x^2\)

2.) \(\displaystyle f(t)=3e^{\sin(2t)}\)

Here we see \(\displaystyle e\) being raised to a power which is a function of \(\displaystyle t\), so we will need to use the chain rule as follows:

\(\displaystyle \frac{d}{dt}\left(e^{u(t)}\right)=e^{u(t)}\frac{du}{dt}\)

Now, we also see that \(\displaystyle u\) is also a function of \(\displaystyle t\), so we need to use the chain rule again. And we also need to use the rule concerning fixed coefficients:

\(\displaystyle \displaystyle \frac{d}{dt}\left(k\cdot f(t)\right)=k\cdot\frac{d}{dt}\left(f(t)\right)\)

So, let's walk though this problem, one step at a time.

\(\displaystyle f(t)=3e^{\sin(2t)}\)

Fixed coefficient:

\(\displaystyle \displaystyle \frac{df}{dt}= 3\frac{d}{dt}\left(e^{\sin(2t)}\right)\)

Exponential and chain rules:

\(\displaystyle \displaystyle \frac{df}{dt}= 3e^{\sin(2t)}\frac{d}{dt}\left(\sin(2t)\right)\)

Sine and chain rules:

\(\displaystyle \displaystyle \frac{df}{dt}= 3e^{\sin(2t)}\cos(2t)\frac{d}{dt}\left(2t\right)\)

Fixed coefficient and power rule:

\(\displaystyle \displaystyle \frac{df}{dt}= 3e^{\sin(2t)}\cos(2t)2\frac{d}{dt}\left(t\right)\)

\(\displaystyle \displaystyle \frac{df}{dt}= 3e^{\sin(2t)}\cos(2t)2(1\cdot t^{1-1})\)

Bring it all together:

\(\displaystyle \displaystyle \frac{df}{dt}= 6\cos(2t)e^{\sin(2t)}\)

3.) \(\displaystyle y=4\cos\left(x^3\right)\)

Coefficient:

\(\displaystyle \displaystyle \frac{dy}{dx}= 4\frac{d}{dx}\left(\cos\left(x^3\right)\right)\)

Cosine and chain rules:

\(\displaystyle \displaystyle \frac{dy}{dx}= 4\left(-\sin\left(x^3\right)\right)\frac{d}{dx} \left(x^3\right)\)

Power rule:

\(\displaystyle \displaystyle \frac{dy}{dx}= -4\sin\left(x^3\right)\left(3x^{3-1}\right)\)

Thus:

\(\displaystyle \displaystyle \frac{dy}{dx}= -12x^2\sin\left(x^3\right)\)

4.) \(\displaystyle \displaystyle f(x)=\frac{e^x}{\sin(x)}\)

This is a quotient, so we can directly apply the quotient rule, and since the arguments of the exponential and sine terms are just \(\displaystyle x\), there's no need for the chain rule.

\(\displaystyle \displaystyle \frac{df}{dx}=\frac{ \sin(x)\dfrac{d}{dx} \left(e^x\right)- e^x\dfrac{d}{dx}\left( \sin(x)\right)}{\sin^2(x)}= \frac{\sin(x)e^x- e^x\cos(x)}{ \sin^2(x)}= \frac{e^x\left( \sin(x)- \cos(x)\right)}{ \sin^2(x)}\)

 

[kmce]

I just realised i put in the first equation wrong, it is

y=x³ ​sin x

 

[MarkFL]

I just realised i put in the first equation wrong, it is

y=x³ ​sin x

Then you could apply the product rule:

\(\displaystyle \displaystyle \frac{dy}{dx}=x^3\frac{d}{dx}\left(\sin(x)\right)+\frac{d}{dx}\left(x^3\right)\sin(x)=\cdots\)

 

[kmce]

I am still slightly confused with how i know what one i should use.This is the first time ive studied math in about 13 years, so simple things are so simple for me anymore

Anyway, so for this one

y=4cos (x³) how did you know it was the chain rule to use. From what ive understood, Product is used when its a function * function (not entirely sure if that is true). So the way ive read this is 4cos * x³​ or am i reading that wrong.

 

[MarkFL]

I am still slightly confused with how i know what one i should use.This is the first time ive studied math in about 13 years, so simple things are so simple for me anymore

Anyway, so for this one

y=4cos (x³) how did you know it was the chain rule to use. From what ive understood, Product is used when its a function * function (not entirely sure if that is true). So the way ive read this is 4cos * x³​ or am i reading that wrong.

The trigonometric functions require an argument, just like any other function, so it would be read as:

y equals 4 times the cosine of x cubed

\(\displaystyle \cos\) by itself is meaningless...we need to know what to take the cosine of, which is this case is the cube of \(\displaystyle x\).

So, we have a constant times a function, where the argument of the function (cosine) is a function itself (cube of \(\displaystyle x\))...and this tells us we will need to use the chain rule.

 

[kmce]

so 3e^sin(2t)

is chain because 2t is a function of sin? How would the 3e come into play the. Sorry this is really confusing me

 

[MarkFL]

so 3e^sin(2t)

is chain because 2t is a function of sin? How would the 3e come into play the. Sorry this is really confusing me

For this one, the chain rule is required because the exponent on \(\displaystyle e\) is a function of \(\displaystyle t\), and then the chain rule is required again because the argument of the sine function in the exponent is also a function of \(\displaystyle t\). It's like we have 3 nested functions. Let:

\(\displaystyle g_1(t)=3e^t\)

\(\displaystyle g_2(t)=\sin(t)\)

\(\displaystyle g_3(t)=2t\)

So, we can write:

\(\displaystyle f(t)=g_1\left(g_2\left(g_3(t)\right)\right)\)

Hence, by the chain rule:

\(\displaystyle f'(t)=g_1'\left(g_2\left(g_3(t)\right)\right) g_2'\left(g_3(t)\right)g_3'(t)\)

Any time you see where one function is nested within another, that's your clue that the chain rule will be needed.

 

[kmce]

In the equation y=3e^sin(2t)

Why does it become 6 cos. I mean, why is the 2 from the cos multiplied by the 3, doesnt the 2 only belong to cos.

[FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]e[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT]dfdt=3esin⁡(2t)cos⁡(2t)2ddt(t)

[FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]e[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT]dfdt=3esin⁡(2t)cos⁡(2t)2(1⋅t1−1)

Bring it all together:

[FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math-italic]e[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT]

 

[MarkFL]

In the equation y=3e^sin(2t)

Why does it become 6 cos. I mean, why is the 2 from the cos multiplied by the 3, doesnt the 2 only belong to cos.

[FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]e[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT]dfdt=3esin⁡(2t)cos⁡(2t)2ddt(t)

[FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]e[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT]dfdt=3esin⁡(2t)cos⁡(2t)2(1⋅t1−1)

Bring it all together:

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If you have:

\(\displaystyle y=a\cdot f(x)\cdot b\cdot g(x)\)

Then by the commutative law of multiplication, you may write:

\(\displaystyle y=a\cdot b\cdot f(x)\cdot g(x)\)

In the example you cited, the 2 was not coupled to the cosine function, it is just a factor at that point, and we may bring like factors together as products. For example:

\(\displaystyle 2x^2\cdot3x^4=6x^6\)

 

[kmce]

ah ok, i hope i remember all this.

So i have 2 moer equations, 1 i think i know what it is, the other i am lost with.

The one i think i know is s=sin(4+t²) I think this is the chain rule, because its a function sin(4+t²) and then has the t² within it.

However the one i am stuck with is y=e^-t. I was initially thinking to change it to ln(-t) and then trying to work it out from there, so ln(1/t) however, I dont actually know if i can do this, since im meant to use either the chain, product and quotient formulas

 

[MarkFL]

ah ok, i hope i remember all this.

So i have 2 moer equations, 1 i think i know what it is, the other i am lost with.

The one i think i know is s=sin(4+t²) I think this is the chain rule, because its a function sin(4+t²) and then has the t² within it.

However the one i am stuck with is y=e^-t. I was initially thinking to change it to ln(-t) and then trying to work it out from there, so ln(1/t) however, I dont actually know if i can do this, since im meant to use either the chain, product and quotient formulas

1.) Okay, we are given:

\(\displaystyle s=\sin\left(4+t^2\right)\)

Yes, we have the function:

\(\displaystyle f(t)=4+t^2\)

as the argument for the sine function...so (using the rule for the sine function and the chain rule) we want:

\(\displaystyle s'=\cos\left(f(t)\right)f'(t)=\,?\)

2.) This time we are given:

\(\displaystyle y=e^{-t}\)

We have the function:

\(\displaystyle f(t)=-t\)

So, using the rule for the exponential function and the chain rule, we have:

\(\displaystyle y=e^{f(t)}f'(t)=\,?\)

 

[kmce]

1.) Okay, we are given:

\(\displaystyle s=\sin\left(4+t^2\right)\)

Yes, we have the function:

\(\displaystyle f(t)=4+t^2\)

as the argument for the sine function...so (using the rule for the sine function and the chain rule) we want:

\(\displaystyle s'=\cos\left(f(t)\right)f'(t)=\,?\)

2.) This time we are given:

\(\displaystyle y=e^{-t}\)

We have the function:

\(\displaystyle f(t)=-t\)

So, using the rule for the exponential function and the chain rule, we have:

\(\displaystyle y=e^{f(t)}f'(t)=\,?\)

Ok so the top one would be

\(\displaystyle cos u\) x \(\displaystyle 4+t2\)


so

\(\displaystyle 2tcos (4+t²​)\)

The second one, i still have no idea, my table says the derivative of an exponential stays the same, so this one i have absolutely no idea what to do with. The only thing i can think is do ln t, then something with 1/t

Edit. Might have worked it out.

So v=e u=t

so it would be e^-t x -t . or something along the lines of that

 

[MarkFL]

Ok so the top one would be

\(\displaystyle cos u\) x \(\displaystyle 4+t2\)


so

\(\displaystyle 2tcos (4+t²​)\)

The second one, i still have no idea, my table says the derivative of an exponential stays the same, so this one i have absolutely no idea what to do with. The only thing i can think is do ln t, then something with 1/t

Edit. Might have worked it out.

So v=e u=t

so it would be e^-t x -t . or something along the lines of that

You are correct that the first one is:

\(\displaystyle s'=2t\cos\left(4+t^2\right)\)

For the second one, we have:

\(\displaystyle \displaystyle y'=e^{-t}\frac{d}{dt}(-t)=-e^{-t}\)

 

[kmce]

Oh i thought it would be something more advanced than that , Thanks for the help

 

[mmm4444bot]

i hope i remember all this.

I used to say the same thing, when I was taking math classes.

At some point, I realized that I needed more than hope. My solution? Extra practice!

I made time to work extra problems, and I continued doing that until I felt confident I had learned what I needed to know. :cool: