# Quick Quotient Rule Question

• Jul 17th 2012, 08:37 AM
AZach
Quick Quotient Rule Question
Hello all, I have been having some trouble grasping the quotient rule. Here's a sample problem, and I'll point out where I get tripped up.

$f(x)=\frac{12.6(4.8^{x})}{x^2}$

1) $f'(x)=\frac{d}{dx}[{12.6(4.8^{x})}]*{x^2}-{12.6(4.8^{x})}*\frac{d}{dx}({x^2})$ all over ${(x^2)^2}$ (I couldn't find the right LaTeX command to divide the entire function by ${(x^2)^2}$)

2) $f'(x)={12.6*(4.8^{x}ln4.8)}]*{x^2}-{12.6(4.8^{x})}*({2x})$ all over ${(x^4)}$

3) I divided ${(x^4)}$ from ${x^2}$ in the numerator, and then combined like terms. $f'(x)={12.6{x^{-2}}*(4.8^{x}ln4.8)}]}-{25.2x(4.8^{x})}$

----I think I see where I went wrong. When I divide ${(x^4)}$ in the denominator, I need to also do it to $\frac{2x}{(x^4)}$?; the book's answer is $f'(x)={12.6(4.8^{x}ln4.8)}}{x^{-2}}-{25.2(4.8^{x})}{x^{-3}}$.

I guess what I'm asking is if my answer in part 3 is correct assuming the book's is just another way to simplify the same problem?
• Jul 17th 2012, 09:15 AM
Reckoner
Re: Quick Quotient Rule Question
Quote:

Originally Posted by AZach
I think I see where I went wrong. When I divide ${(x^4)}$ in the denominator, I need to also do it to $\frac{2x}{(x^4)}$?; the book's answer is $f'(x)={12.6(4.8^{x}ln4.8)}}{x^{-2}}-{25.2(4.8^{x})}{x^{-3}}$.

I guess what I'm asking is if my answer in part 3 is correct assuming the book's is just another way to simplify the same problem?

Your answer in part 3 is incorrect, for the reason that you stated. In general, you can cancel factors of the numerator and denominator. But if you want to cancel a factor from an individual term, that is not allowed. You must cancel from every term in the numerator.

Continuing from step 2, we have

$f'(x) = \frac{12.6x^2\left(4.8^x\ln4.8\right) - 25.2x\left(4.8^x\right)}{x^4}$

$f'(x) = \frac{x^4\left[12.6x^{-2}\left(4.8^x\ln4.8\right) - 25.2x^{-3}\left(4.8^x\right)\right]}{x^4}$

$f'(x) = 12.6x^{-2}\left(4.8^x\ln4.8\right) - 25.2x^{-3}\left(4.8^x\right)$