Question

i need helping figuring out the tangent line im still stuck!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! this was the original problem
f(x)= [1/2x][lnx^4] then f'(x)= 1/2 lnx^4 + 4 and given points are (-1,0). and NOW that the derivative is taken, i need to plug in the value for x which is -1 and solve for m and then plug it into the point slope formula. but im stuck when it comes to solving for m. on my calculator i got and error, non real result. ;(

Answer 1

oh is that exponent on the x
or ln(x)

Answer 2

To begin with I think your derivative is wrong

Answer 3

original prooblem is 1/2x ln(x^4)

Answer 4

Did you use quotient or product rule?

Answer 5

the derivative is 1/2ln(x)^4+4

Answer 6

im stuck at finding the slope!!!!

Answer 7

\[f(x)=\frac{1}{2}x 4\ln(x)=2xln(x)\]
\[f'(x)=2\ln(x)+2x \frac{1}{x}=2\ln(x)+2\]

Answer 8

i plugged in -1 for x, like im supposed to use point slope formula but im stuck at finding for m

Answer 9

oh so i have the wrong derivative. its really what u just told me ?

Answer 10

...yeah I think so let me solve and see what I get..but first is the question: x /2 or 1 / 2x

Answer 11

its 1/2x

Answer 12

ok

Answer 13

so is it x/2 or 1/(2x) ?

Answer 14

1/(2x)

Answer 15

ok so what i have is not right

Answer 16

unless it is x/2 which it is not

Answer 17

its one half then x then ln of x raised to the 4th

Answer 18

\[y=\frac{1}{2x}\ln(x^4)=\frac{1}{2}(\frac{1}{x}\ln(x^4))\]
\[y'=\frac{1}{2}(\frac{-1}{x^2}\ln(x^4)+\frac{1}{x} \frac{4x^3}{x^4})\]

Answer 19

\[y'=\frac{1}{2}(-\frac{1}{x^2}\ln(x^4)+\frac{4}{x^2})=\frac{1}{2x^2}(-\ln(x^4)+4)\]

Answer 20

\[\ \frac{1-lnx}{2x ^{2}} \]

Using quotient rule\[f \prime = \ \frac{1}{2} ( \frac{lnx}{x} ) = \ \frac{1}{2} (\frac{1- lnx}{x ^{2}} )\]

Answer 21

\[y'=\frac{1}{2(1)}(-\ln(1)+4)=2\]

Answer 22

ok i guess i don't know what the function is lol

Answer 23

no you're right I just did the quotient rule

Answer 24

oh this is what you got for your derivative
\[\ \frac{1-lnx}{2x ^{2}} \]
i thought u were meaning to be the original function
ok cool

Answer 25

i see

Answer 26

yeah aren't we suppose to plug in -1 for x

Answer 27

so the slope of tangent line at x=-1 is 2

Answer 28

ln(-1) is invalid

Answer 29

ln(x^4) ?

Answer 30

yes

Answer 31

oh okay I brought the 4 down as a coefficient