Mastering the Chain Rule with Fractions for Calculus Students

[TheAkuma]

Okay, I know how to differentiate regular functions. But when it comes to fractions, I'm hopeless. This may be an extremely simple one to some, here is the function; "1/4x-7"
I have to differentiate that using the chain rule.

I think that u=4x-7, but I am not sure. As i said, I am horrible when it comes to fractions in the chain rule.

 

[HallsofIvy]

Okay, I know how to differentiate regular functions. But when it comes to fractions, I'm hopeless. This may be an extremely simple one to some, here is the function; "1/4x-7"
I have to differentiate that using the chain rule.

I think that u=4x-7, but I am not sure. As i said, I am horrible when it comes to fractions in the chain rule.

First, what exactly is your function? Is it (1/4)x- 7 or 1/(4x)- 7 or 1/(4x- 7)?

Since you say "I think that u=4x-7", I assume it is 1/(4x- 7). As for what "u" is, that's your choice. TRY something and see if it works. If you choose u= 4x- 7 then 1/(4x-7)= 1/u= u^-1. Can you differentiate that? And you certainly ought to be able to differentiate u= 4x-7 (the derivative of a linear function is just its slope). Finally, the chain rule says
[tex]\frac{df}{dx}= \frac{df}{du}\frac{du}{dx}[/tex].

 

[TheAkuma]

First, what exactly is your function? Is it (1/4)x- 7 or 1/(4x)- 7 or 1/(4x- 7)?

Since you say "I think that u=4x-7", I assume it is 1/(4x- 7). As for what "u" is, that's your choice. TRY something and see if it works. If you choose u= 4x- 7 then 1/(4x-7)= 1/u= u^-1. Can you differentiate that? And you certainly ought to be able to differentiate u= 4x-7 (the derivative of a linear function is just its slope). Finally, the chain rule says
[tex]\frac{df}{dx}= \frac{df}{du}\frac{du}{dx}[/tex].

well the function is actually 1/4x-7. there is no brackets so i naturally assumed u=4x-7. I get it how the answer is -4/(4x-7)² but why is it squared? i was thinking 1/u X -4 where u=4x-7. What I want to know now is why is the u squared?

 

[TheAkuma]

Ohh! when there is a power of at the bottom of the fraction, does it go up? like from to the power of one it'll go up to the power of 2?

 

[Mark44]

I'm guessing that your function is written in your text or worksheet like this:
[tex]\frac{1}{4x - 7}[/tex]

When you write it on a single line, you have to put parentheses around the terms in the denominator. IOW, like this: 1/(4x - 7).

The way you wrote it, without parentheses, would be interpreted like this:
[tex]\frac{1}{4} x - 7[/tex]

BTW, why did you post this under Precalculus Mathematics? This is obviously a calculus problem.

 

[HallsofIvy]

Ohh! when there is a power of at the bottom of the fraction, does it go up? like from to the power of one it'll go up to the power of 2?

[tex]\frac{1}{x^2}= x^{-2}[/tex] and the derivative of [itex]x^n[/itex] is [itex]nx^{n-1}[/itex] where n is any number [As GibZ pointed out- any number except -1!]. If n= -2 what does that give you?

(You could also do that problem using the "quotient rule": the derivative of u(x)/v(x)= (u'v- uv')/v². If u=1 and v= x², what does that give you?

 

[TheAkuma]

I'm guessing that your function is written in your text or worksheet like this:
[tex]\frac{1}{4x - 7}[/tex]

When you write it on a single line, you have to put parentheses around the terms in the denominator. IOW, like this: 1/(4x - 7).

The way you wrote it, without parentheses, would be interpreted like this:
[tex]\frac{1}{4} x - 7[/tex]

BTW, why did you post this under Precalculus Mathematics? This is obviously a calculus problem.

Well my teacher told me that this is sort of like the introduction to calculus.

 

[Gib Z]

[tex]\frac{1}{x^2}= x^{-2}[/tex] and the derivative of [itex]x^n[/itex] is [itex]nx^{n-1}[/itex] where n is any number. If n= -2 what does that give you?

Careful! Not if n=0 !. =]

TheAkuma: Introduction or not, please post under the calculus section. If helps the homework helpers out if they know what kind of things to expect to help people with.