Derivatives

[jkeatin]

1. The problem statement, all variables and given/known data
Use the differential dy or L to approximate cubed root 1.03


2. Relevant equations
L(x)= f(a) +f'(a)(x-a)


3. The attempt at a solution
I have no clue how to start so that would be the most help, do i just find the derivative of cubed root 1.03 first?

[Dick]

You find the derivative of a^(1/3) at a=1 first. That's f(a). Then x=1.03. So x-a=0.03. Does that help?

[jkeatin]

so
L(x)= (a)^1/3+ 1/3(a)^-1/3(x-a)
substitue 1 for a only in the functions right

[gamesguru]

Your only equation was,
L(x)= f(a) +f'(a)(x-a).
This gives a good approximation to values of a function near x=a. So take the derivative of x^1/3.
f'(x)=d/dx x^1/3=1/3x^-2/3.
Now plug in a=1 and you should be good and you have an approximation near x=1.
Hope that helps.

[jkeatin]

is it l(x)= 1/3+1/3(0.03)

[gamesguru]

No, that's wrong. I'll work it out for you so hopefully you learn the process.
My other post said that f'(x)=1/3x^-2/3, I just used the exponent rule.
In the linearization, a is called the center and is where the linearization is most accurate. If we try to use it to find the cube root of 1000, it won't be very accurate.
The linearization around a=1 is,
L(x)=1^1/3+1/3(x-1)=1+1/3(x-1).
Now you just need to plug in x=1.03 to get the approximate value for the cube root of 1.03.

[jkeatin]

when you substitue 1 in f'(x) that equals 1^1/3?

[jkeatin]

or 1/3

[gamesguru]

Yes...
f'(x)=1/3*x^(-2/3)
f'(1)=1/3*1^(-2/3)=1/3*1=1/3.

[jkeatin]

sorry first time using this online thing i not used to way things are written, thanks for all your help.