Question

The twice-differentiable function f is defined for all real numbers and satisfies the following conditions: f(0)= 4, f'(0)= -3, and f''(0)= 2 a) the function is given by g(x)= a ln(f(x)) for all real numbers, where a is a constant.Find g'(0) and g''(0) in terms of a. Show the work that leads to your answers b) the function h is given by h(x) =sec(kx)f(x) for all real numbers, where k is a constant. Find h'(x) and write an equation for the line tangent to the graph of h at x=0

Answer 1

to start off with

g'(x) = af'(x)/f(x)

so g(0) = -3a/4

Answer 2

sorry g'(0) = -3a/4

Answer 3

Do you know how to differentiate?

Answer 4

my working goes like this.........

g(x)= a ln[f(x)]

the derivative of lnx is 1/x

so the derivative of ln[f(x)] should be [1/f(x)].f'(x)

hence g'(x) = af'(x)/f(x)

Answer 5

let me know if you think this is wrong

Answer 6

g''(x) is a little more complex

Answer 7

I have an answer for this as well but i'll let dmendez carry on from here

Answer 8

Thanks for your help, I didn't understand what the a in the first given function meant but now I know how to find the derivative of F(x) to find g'(x)

Answer 9

I'm having trouble finding g"(x) though

Answer 10

yeah its tricky

you have to use the quotient rule

g'(u/v) = (u'v-v'u)/v^2

where u = af'(x) and v = f(x)

Answer 11

you should get g''(0) = -a/16

Answer 12

give it a try and see how you go while i work on the second part

Answer 13

okay, thanks

Answer 14

okay so, the derivative of g(x) would be ((a f'(x))/(f(x))

Answer 15

yes, that's right, but what did you get for the 2nd derivative of g(x) ?

Answer 16

yeah, would it be (a f"(x)=(a f'(x)^2)/(f(x))+f(x) g"(x))?

Answer 17

no its........

g'' = (af''f -af'^2)/f^2

i've left out the x's for simplicity

Answer 18

is (a(-f'(x)^2)+f(x)f"(x)))/f(x)^2=g"(x)) correct?

Answer 19

yes thats it!

Answer 20

Cool, Thank you so much for your help!

Answer 21

no problem. i'll let you do the rest.

if you need help with the 2nd part let me know

Answer 22

thanks, the second part I understand:)