Question

find the derivative of f(x)= e^2x (x^2 + 5^x)

Answer 1

\[\Large f(x)\quad=\quad e^{2x}\left(x^2+5^x\right)\]So looks like we would start by applying the product rule.
What part is giving you trouble?
Forget how to take the derivative of that exponential part? :O 5^x

Answer 2

well ive been trying to apply the product rule but im not getting the right answer :c
so far i have \[f'(x)= e ^{2x}[2x+(\ln5\times5^x)] + (x^2 + 5^x)(2e ^{2x})\]
that right?

Answer 3

yes looks good so far! :)

Answer 4

i guess now i just have no idea how to simplify that.

Answer 5

Hmm well they both have the same exponential right?
let's factor that out.
Maybe it will make things easier.\[\Large f'(x)\quad=\quad \color{orangered}{e^{2x}}\left[2x+5^x \ln5+x^2+5^x\right]\]

Answer 6

Woops I missed a 2, sec..

Answer 7

\[\Large f'(x)\quad=\quad \color{orangered}{e^{2x}}\left[2x+5^x \ln5+2(x^2+5^x)\right]\]

Answer 8

Hmm I dunno.. it doesn't really seem like you need to simplify this +_+

Answer 9

You could factor a 2x out of each of the x terms I guess...
And a 5^x out of each of the 5^x terms.\[\Large f'(x)\quad=\quad e^{2x}\left[2x(1+x)+5^x(\ln5+1)\right]\]

Answer 10

It doesn't clean up much better than that D:

Answer 11

the answer is \[e ^{2x}[2x^2 + 2x + (\ln5 +2)5^x]\]

Answer 12

hmm

Answer 13

Oh i Simplified too far :) lol
and i forgot that darn 2 again!!! grr

Answer 14

So go from here:\[\Large f'(x)\quad=\quad \color{orangered}{e^{2x}}\left(2x+5^x \ln5\right)+(x^2+5^x)2\color{orangered}{e^{2x}}\]To here:
\[\Large f'(x)\quad=\quad \color{orangered}{e^{2x}}\left[2x+5^x \ln5+2(x^2+5^x)\right]\]

Answer 15

That step make sense?

Answer 16

I factored the orange part out of each term.

Answer 17

yes that makes sense so far!

Answer 18

That's pretty close to the answer you have.
Just gotta move a few things around :o

Answer 19

im honestly just not understanding the part how it simplified to (ln5 + 2)5^x
sorry ):

Answer 20

\[\Large f'(x)\quad=\quad e^{2x}\left[2x+5^x \ln5+\color{royalblue}{2(x^2+5^x)}\right]\]Let's distribute the 2 to each term in this set of brackets.\[\Large f'(x)\quad=\quad e^{2x}\left[2x+5^x \ln5+\color{royalblue}{2x^2+2\cdot5^x}\right]\]

Answer 21

From there let's rearrange some stuff,\[\Large f'(x)\quad=\quad e^{2x}\left[2x+2x^2+5^x \ln5+2\cdot5^x\right]\]

Answer 22

OOOH okay i understand (: thank you zep!

Answer 23

yay team \c:/