# Thread: [SOLVED] Really quick arithmetic question!!

1. ## [SOLVED] Really quick arithmetic question!!

Hi guys, I'm working my calculus HW right now and this one problem for some reason really confused me and for the life of me I cannot wrap my head around it. My question is not so much about the concept, as it is about the quotient rule and how it is applied.

Given $\displaystyle f'(x)=\frac{(e^{4x+5})(2)-(2x+3)(e^{4x+5})(4)}{e^{(4x+5)^2}}$

What would be the best way to go about simplifying this fraction? I'm afraid I've kind of forgotten rules for cancelling terms within expressions like this one. If someone could lay it out for me, I'd really appreciate it! Thanks

2. Originally Posted by rust1477
Hi guys, I'm working my calculus HW right now and this one problem for some reason really confused me and for the life of me I cannot wrap my head around it. My question is not so much about the concept, as it is about the quotient rule and how it is applied.

Given $\displaystyle f'(x)=\frac{(e^{4x+5})(2)-(2x+3)(e^{4x+5})(4)}{e^{(4x+5)^2}}$

What would be the best way to go about simplifying this fraction? I'm afraid I've kind of forgotten rules for cancelling terms within expressions like this one. If someone could lay it out for me, I'd really appreciate it! Thanks
factor out $\displaystyle 2e^{4x+5}$ from both terms in the numerator, then simplify.

btw ... your denominator should be $\displaystyle e^{2(4x+5)}$

3. Remember that when you have two common bases raised to a power and you divide them you subtract the exponents.

You'll get $\displaystyle e^{-(4x+5)}$, though it won't do any good to simplify evaluating it. The quotient rule is just a disguised form of the power rule.

4. Originally Posted by skeeter
factor out $\displaystyle 2e^{4x+5}$ from both terms in the numerator, then simplify.

btw ... your denominator should be $\displaystyle e^{2(4x+5)}$

Thanks very much!

But, can you elaborate on those steps? The answers I have to choose from all have $\displaystyle e^{4x+5}$ in the denominator, which leads me to believe one of the terms in the numerator got cancelled by it, can you clear this up for me?

5. Because your denominator is e^(4x + 5) * e^(4x + 5). Just like x^4 * x^6 = x^10, you add the exponents, you don't multiply them.

Edit: Guess we don't need to address that anymore. =p

Recall that when taking derivatives and integrals of exponentials, we always still have e^u. So when you derive (which you haven't yet, only simplified), your u'v - v'u is going to have two e^u again, and you'll simplify your v^2 denominator yet again.

Or is it that you have your derivative and you were just simplifying it to get it to look like the answers in the book? If that's the case, know that you should certainly have an e^(4x+5) when you simplify that equation.

6. $\displaystyle f'(x)=\frac{(e^{4x+5})(2)-(2x+3)(e^{4x+5})(4)}{e^{2(4x+5)}}$

$\displaystyle f'(x)=\frac{2e^{4x+5}[1 - 2(2x+3)]}{e^{2(4x+5)}}$

$\displaystyle f'(x)= -\frac{2(4x+5)}{e^{4x+5}}$

7. Perfect, Thanks guys!