Question

If f(x) = 2 x^{4 x}, find f '( 2 ).

Answer 1

could you find the derivative f'(x) first ?

Answer 2

that's the thing I tried, but its wrong

Answer 3

can you show your work ?
i'll try and find the error

Answer 4

2*(4x)x^(4x-1) this isn't right thought

Answer 5

I also tries chain rule. but what would be g(x)??

Answer 6

do u know logarithmic differentiation ?

Answer 7

\[\Large f(x)=2x^{4x}\]
or can be written as
\[\Large y=2x^{4x}\]
take natural log of both sides
\[\Large \ln(y)=\ln(2x^{(4x)})\]

Answer 8

now expand the right side little bit
using property of log
\[\Large \ln(y) =\ln(2) +\ln(x^{4x})\]
\[\Large \ln(y)=\ln(2)+4xln(x)\]
differentiating both sides with respect to x
\[\Large \frac{d(\ln(y)}{dx}=\frac{d(\ln(2)}{dx}+\frac{d}{dx} (4x*(\ln(x))\]
since derivative of constant is zero so it becomes
\[\Large \frac{d(\ln(y)}{dx}=\frac{d}{dx} (4x*(\ln(x))\]

just apply product rule on right side.
let me know result after applying product rule.

Answer 9

ok but wouldn't the left side be ln'(y)

Answer 10

it will be
but
\[\Large \ln'(y)=\frac{1}{y}\frac{dy}{dx}\]

Answer 11

for right side: 4xln(x) + 4

Answer 12

thats correct

Answer 13

the left side as i wrote above is
\[\Large \frac{1}{y}\frac{dy}{dx}\]

Answer 14

writing both together
\[\Large \frac{1}{y}\frac{dy}{dx}=4\ln(x)+4\]

Answer 15

but \[\Large y=2x^{4x}\]

Answer 16

isn't it 4xln(x) not 4ln(x)

Answer 17

no it should be 4ln(x)+4 after product rule because
\[\Large \frac{d}{dx}(4x)=4\]

Answer 18

let me write product rule expansion
\[\Large \frac{d}{dx}(4x*\ln(x)=\frac{d}{dx}(4x)\ln(x)+ 4x \frac{d}{dx}(\ln(x))\]

Answer 19

got it ?

Answer 20

going back to previous result
\[\Large \frac{dy}{dx}=y(4\ln(x+4)\]
but
\[\Large y=2x^{4x}\]
replace the value of y in the above expression. let me know the result.

Answer 21

@jolee did you get it ?

Answer 22

im supposed to find y' though not d/dx

Answer 23

put u put it all on one page

Answer 24

please help fast!!!

Answer 25

these all notations are same.
\[\large \frac{dy}{dx}=y'=f(x)\]

Answer 26

yes but f'(x)

Answer 27

\[\large \frac{dy}{dx}=y'=f'(x)\]

Answer 28

after replacement of y you should have
\[\Large \frac{dy}{dx}=f'(x)=2x^{4x}(4\ln(x)+4)\]

Answer 29

just substitute x=2 in the above expression and use calculator to get f'(2) .

Answer 30

are u certain this is correct?

Answer 31

yes it is .
you can simply it little more if you want.

Answer 32

is the answer 181708?

Answer 33

actually 443848

Answer 34

?

Answer 35

no did you convert your calculator to radians ?

Answer 36

yes im in radian mode. is this wrong?

Answer 37

why would that matter?

Answer 38

all calculations in calculus are done in radians.
try again because i am getting 3467.566

Answer 39

but why in radians? srry.

Answer 40

just for the sake of convience.

Answer 41

I keep getting 443848. first from ln part, 6.77 times 4^8

Answer 42

2*2^8*6.77

Answer 43

ahh thank you. do u also know physics by chance?

Answer 44

by chance i do know. :)

Answer 45

ill ask a question in physics section then!