SOLUTION: if f(x) = {{{ln(5x^2 +1)^3}}}, then the derivative of f evaluated at x=1 equals: a) 2 b) 4 c) 5 d) 6 My guess is D (6). I graphed the function. I used my table command

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: if f(x) = {{{ln(5x^2 +1)^3}}}, then the derivative of f evaluated at x=1 equals: a) 2 b) 4 c) 5 d) 6 My guess is D (6). I graphed the function. I used my table command       Log On


   

Question 281944: if f(x) = ln%285x%5E2+%2B1%29%5E3, then the derivative of f evaluated at x=1 equals:
a) 2 b) 4 c) 5 d) 6
My guess is D (6). I graphed the function. I used my table command with the change in x being .1 Then I found the value of the function at 1.9 and 2.1 Since the derivative represents the slope at a particular x-value, I found the slope around 2. The slope was 5.9595, which is close to 6.
I tried to take the derivative of this function, but I got lost. Here's what I tried:
ln%285x%5E2%2B1%29%5E3
=1%2F%285x%5E2+%2B1%29%5E3+%2A+3%285x%5E2%2B1%29%5E2%2810x%29
=30x%2F%285x%5E2+%2B1%29 then the functional value at 1 is 30/6 = 5.
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
>if f(x) = ln%285x%5E2+%2B1%29%5E3, then the derivative of f evaluated at x=1 equals:
>a) 2 b) 4 c) 5 d) 6
>My guess is D (6). I graphed the function. I used my table command with the >change in x being .1 Then I found the value of the function at 1.9 and 2.1
Why 1.9 and 2.1?
> Since the derivative represents the slope at a particular x-value,
True.
> I found the slope around 2.
Again, why a number near 2?
> The slope was 5.9595, which is close to 6.
You found the slope with the calculator? However, you're doing it there is an error. (I'll show you the actual slope at 2 later.)
> I tried to take the derivative of this function, but I got lost.
No you didn't!
> Here's what I tried:
>f%28x%29+=+ln%285x%5E2%2B1%29%5E3
>f'(x) = 1%2F%285x%5E2+%2B1%29%5E3+%2A+3%285x%5E2%2B1%29%5E2%2810x%29
>f'(x) = 30x%2F%285x%5E2+%2B1%29
>then the functional value at 1 is 30/6 = 5.
This is all totally correct! (Although I'd word it: "The value of the (first) derivative at 1 is 30/6 = 5") The answer to your problem is 5.
Again, I don't get your fixation on number near 2. They have nothing to do with the problem at hand. But FWIW,
f'(2) = which is approximately: 2.85714286