Differentiating fractions

[joshmeyer]

$$(8x^2-5)/(4x^3+1)$$I understand how to differentiate fractions, but what I don't understand is which number I use first. 5-4 is very different from 4-5, but addition doesn't have similar problems

would it be
$$((8x^2-5)(12x^2)-(16x)(4x^3+1))/(4x^3+1)^2$$or
$$((16x)(4x^3+1)-(8x^2-5)(12x^2))/(4x^3+1)^2$$and why?

 

[Deleted member 4993]

$$(8x^2-5)/(4x^3+1)$$I understand how to differentiate fractions, but what I don't understand is which number I use first. 5-4 is very different from 4-5, but addition doesn't have similar problems

would it be
$$((8x^2-5)(12x^2)-(16x)(4x^3+1))/(4x^3+1)^2$$or
$$((16x)(4x^3+1)-(8x^2-5)(12x^2))/(4x^3+1)^2$$and why?

If f(x) = $\displaystyle \frac{N(x)}{D(x)}$ .........................................then

$\displaystyle \frac{d}{dx}f(x) = f'(x) = \frac{N'(x) * D(x) - D'(x) * N(x)}{\left(D(x)\right)^2}$

This is sometimes called the "Quotient Rule" of differentiation. Proof of this should be available in any calculus book.

Can you prove the "product rule"?

 

[joshmeyer]

If f(x) = $\displaystyle \frac{N(x)}{D(x)}$ .........................................then

$\displaystyle \frac{d}{dx}f(x) = f'(x) = \frac{N'(x) * D(x) - D'(x) * N(x)}{\left(D(x)\right)^2}$

This is sometimes called the "Quotient Rule" of differentiation. Proof of this should be available in any calculus book.

Can you prove the "product rule"?

so the first prime value is always the numerator? and proving the product rule is using the x+h formula right?

 

[topsquark]

Another way, if you have a bad memory like me. You can rewrite this in terms of the product rule and use the chain rule:
$$\dfrac{d}{dx} \dfrac{N(x)}{D(x)} = \dfrac{d}{dx} \left ( N(x) (D(x)^{-1}) \right ) = \dfrac{dN}{dx} (D(x))^{-1} + N(x) \left ( -1 (D(x)^{-2}) \dfrac{dD}{dx} \right )$$
-Dan

 

[Dr.Peterson]

$$(8x^2-5)/(4x^3+1)$$I understand how to differentiate fractions, but what I don't understand is which number I use first. 5-4 is very different from 4-5, but addition doesn't have similar problems

would it be
$$((8x^2-5)(12x^2)-(16x)(4x^3+1))/(4x^3+1)^2$$or
$$((16x)(4x^3+1)-(8x^2-5)(12x^2))/(4x^3+1)^2$$and why?

The way I remember it is to differentiate the top first: (N/D)' = (N'D - ND')/D^2.

In order to remember this better, I also remember the product rule in the same way, differentiating the first factor first, even though in this case the order doesn't matter: (AB)' = A'B + AB'.

But if I haven't done it lately, I do exactly what @topsquark said. In fact, sometimes that's a better way in the first place.

 

[Deleted member 4993]

Another way, if you have a bad memory like me. You can rewrite this in terms of the product rule and use the chain rule:
$$\dfrac{d}{dx} \dfrac{N(x)}{D(x)} = \dfrac{d}{dx} \left ( N(x) (D(x)^{-1}) \right ) = \dfrac{dN}{dx} (D(x))^{-1} + N(x) \left ( -1 (D(x)^{-2}) \dfrac{dD}{dx} \right )$$
-Dan

That is why exactly I had asked the poster about product rule. Whenever I have to use quotient rule - I re-derive it from the "product rule" 99.237% times.

 

[joshmeyer]

Another way, if you have a bad memory like me. You can rewrite this in terms of the product rule and use the chain rule:
$$\dfrac{d}{dx} \dfrac{N(x)}{D(x)} = \dfrac{d}{dx} \left ( N(x) (D(x)^{-1}) \right ) = \dfrac{dN}{dx} (D(x))^{-1} + N(x) \left ( -1 (D(x)^{-2}) \dfrac{dD}{dx} \right )$$
-Dan

would you mind expanding this using my problem? I'm having trouble plugging this in.

 

[Deleted member 4993]

would you mind expanding this using my problem? I'm having trouble plugging this in.

In your case, please tell us:

What is N(x) ? ......................... N(x) = ?

What is N'(x) ? ......................... N'(x) = ?

What is D(x) ? ......................... D(x) = ?

What is D'(x) ? ......................... D'(x) = ?

What is D²(x) ? ......................... D²(x) = ?

 

[Dr.Peterson]

would you mind expanding this using my problem? I'm having trouble plugging this in.

You're asking about the use of the product rule, rather than the quotient rule, which you already got right:

$$((16x)(4x^3+1)-(8x^2-5)(12x^2))/(4x^3+1)^2$$

Here is the start of the product rule approach:

$$f(x) = \frac{8x^2−5}{4x^3+1} = (8x^2−5)(4x^3+1)^{-1}$$
So, you will be differentiating $(8x^2−5)$ and $(4x^3+1)^{-1}$. Give that a try.

 

[Steven G]

I always wonder why Calculus books make the product rule and quotient rule look so differently.
The product rule for $\displaystyle f(x)g(x)\ is\ f(x)g'(x) + f'(x)g(x)$.
The numerator of the quotient rule is exactly the same as the product rule except that the addition sign is a subtraction sign.
To make the quotient rule look like the product rule I state that the derivative of
$\displaystyle \dfrac{f(x)}{g(x)}\ is\ \dfrac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$

 

[Harry_the_cat]

If $\displaystyle y=\frac{u}{v}$ then $\displaystyle y'=\frac{u'v - uv'}{v^2}$.

The way I remember it is by looking at the "powers" of $\displaystyle v$, although they are not really "powers".
Reading from L to R, the first v has nothing above it, the second v has a 1 (really an ' ) and the last v has a 2.
And remembering the minus sign on the top.
Works for me!

 

[joshmeyer]

Thanks everyone! I understand how it works now.