Question

derivative f(x)=3^(4x^5+x-8)*log[2](6x+3)

Answer 1

\[y=3^{4x^5+x-8} \cdot \log_2(6x+3) ?\]

Answer 2

yep thats the one!

Answer 3

so we have a product so we need the product rule
y'=f'g+fg'
right?

Answer 4

Now lets look at f
\[f=3^{4x^5+x-8} ; g=\log_2(6x+3)\]

Answer 5

oops lets look at f first ok?

Answer 6

Do you know how to find derivative of that?

Answer 7

no

Answer 8

\[f=3^{p(x)}\]
to find derivative of this take ln( ) of both sides

Answer 9

\[\ln(f)=\ln(3^{p(x)})\]

Answer 10

\[\ln(f)=p(x) \ln(3)\]

Answer 11

Take derivative of both sides

Answer 12

soryy but could u please explane me a little bit more! i am really bad in derivatives!

Answer 13

f and p are both functions of x

Answer 14

(ln(f))'=?

Answer 15

f'/f right?

Answer 16

f=3^(4x^5+x−8)'/3^(4x^5+x−8) thats right

Answer 17

no no

Answer 18

i see i am really bad at this :)

Answer 19

\[\frac{f'}{f}=\ln(3) p'(x)\] <---i'm trying to find f'

Answer 20

we need to solve this for f'

Answer 21

how do we do it?

Answer 22

no idea! :) thats the one i dont know how to do!

Answer 23

you can solve for f' just multiply f on both sides and therefore f' is isolated

Answer 24

\[\frac{d}{dx}b^x=b^x\ln(b)\] so by the chain rule
\[\frac{d}{dx}3^{4x^5+x-8}=\ln(3) 3^{4x^5+x-8}\times (20x+1)\]

Answer 25

\[\frac{d}{dx}\log_b(x)=\frac{1}{x\ln(b)}\]so again by the chain rule
\[\frac{d}{dx}\log_2(6x+3)=\frac{1}{(6x+3)\ln(2)}\times 6\]

Answer 26

clean it up to get
\[\frac{d}{dx}\log_2(6x+3)=\frac{3}{(2x+1)\ln(2)}\] and then use the product rule to get your answer

Answer 27

\[(fg)'=f'g+gf'\] plug in the functions and the derivatives, don't simplify

Answer 28

thank u!