1. ## Supposedly easy derivative:

Hello all,

I'm in charge of finding the derivative of $\displaystyle 1 \over 4x-3$.

I first used the quotient rule, which came out to be

$\displaystyle 0 * (4x-3) - ((4x-3) * 1) \over (4x-3)^2$

and simplified to

$\displaystyle - 4x-3 \over (4x-3)^2$

However, this isn't complete. I understand that I need to use the chain rule on the numerator next, according to Wolfram-Alpha, but I don't understand why I need the chain rule since there's only 1 function. Can anyone shed some light on this?

2. Use the law of exponents that says $\displaystyle \dfrac{1}{a} = a^{-1}$ to rewrite $\displaystyle \dfrac{1}{4x-3} = (4x-3)^{-1}$

You can the use the chain rule

3. Oh wow! Haha.

Now that you show that, it's pretty easy to see. I'd imagine I should do this for all fractions with 1 in the numerator correct?

It makes complete sense:

$\displaystyle (4x-3)^{-1}$

$\displaystyle (4x-3)^{-1} = (4x-3)^{-2} * 4$

$\displaystyle 4 \over (4x-3)^2$

EDIT: I should say, a constant in the numerator

4. Yes, but bear in mind that it will not be affected by the exponent. That is $\displaystyle \dfrac{2}{x-1} = 2(x-1)^{-1} \neq (2(x-1))^{-1}$

5. Originally Posted by sinx
Hello all,

I'm in charge of finding the derivative of $\displaystyle 1 \over 4x-3$.

I first used the quotient rule, which came out to be

$\displaystyle 0 * (4x-3) - ((4x-3) * 1) \over (4x-3)^2$
No, that's not correct. The quotient rule says that
$\displaystyle \left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}$
Here f(x)= 1 so f'(x)= 0, g(x)= 4x- 3 so that g'= 4.
$\displaystyle \frac{0(4x-3)- (4)(1)}{(4x- 3)^2}$

and simplified to

$\displaystyle - 4x-3 \over (4x-3)^2$

However, this isn't complete. I understand that I need to use the chain rule on the numerator next, according to Wolfram-Alpha, but I don't understand why I need the chain rule since there's only 1 function. Can anyone shed some light on this?