Question

Calculus:

Find the derivative of the following function-
-4e^5x/7x+2

Answer 1

Use the quotient rule.

Answer 2

\[(\frac{f}{g})'=\frac{gf'-f'g}{g^2}\]

Answer 3

put
\[f(x)=-5e^{5x}\]
\[f'(x)=-25e^{5x}\]
\[g(x)=7x+2\]
\[g'(x)=7\]

Answer 4

put em in and out pops the answer.

Answer 5

except of course my formula is wrong!

Answer 6

Should be:\[(\frac{f}{g})=\frac{f'g-g'f}{g^2}\]

Answer 7

\[\frac{gf'-fg'}{g^2}\]

Answer 8

what bendt said! sorry

Answer 9

What I'm having trouble with is simplifying the answer at the end.

Answer 10

What do you get?

Answer 11

\[\frac{-5e^{5x}\times 7 - (7x+2)\times -25e^{5x}}{(7x+2)^2}\]

Answer 12

numerator is
\[-35e^{5x}+175xe^{5x}+40e{5x}\]
\[=175xe{5x}+5e^{5x}\]
\[=e^{5x}(175x+5)\]
\[=5e^{5x}(35x+1)\]unless i made an algebra mistake which is always possible. bendt?

Answer 13

Can we cheat and use Wolfram Alpha to check?

Answer 14

ok i made a mistake on the very first line.
\[f(x)=-4e^{5x}\]
\[f'(x)=-20e^{5x}\]

Answer 15

i wrote -25 and that was a mistake

Answer 16

So do you start off by factoring out the e^5x?

Answer 17

Oh yeah.

Answer 18

i think my last answer is correct though i will check it since i messed up

Answer 19

yes it factors out of the whole thing

Answer 20

yea i screwed up. numerator is
\[-4e^{5x}(35x+3)\]

Answer 21

sorry

Answer 22

Thank you two very much for your help. This class is very difficult to me :(

Answer 23

sorry for my error. that -25 threw me off. use -20 and it should be ok

Answer 24

I'm confused, I think the expression should be:\[\frac{(-20e^{5x})\times(7x+2)-7\times(-4e^{5x})}{(7x-2)^2}\]

Answer 25

Yes, I think satellite73 figured out the problem using -5e^5x instead of -4e^5x by mistake.

Answer 26

Ah, ok, as long as you know what your doing now.