Differentiation and tangent line

  • #1

Homework Statement

f(x) = KxL

K= 1.78
L= -1.39
Problem 1: Find f'(x).
v= 0.89
w= 0.5
"v" and "w" are two points located on the x-axis.
Problem 2: Calculate f'(v).
Problem 3: Find the equation of the tangent line of f(x) over the point "w". The equation should be in this form y=Ax+B

Problem 1:This is how I worked out the differentiation:
f(x) = e(ln 1.78)x-1.39
Setting u=x-1.39,
f'(x) = (Ku)'(u) * u'

(Ku)'(u) = (eu ln K)' = ln K * eu ln K = ln K * Ku
u' = LxL-1 = -1.39x-2.39

f'(x) = ln 1.78(1.78x-1.39) * -1.39x-2.39

Is this correct?
Regarding the other problems, do I need f'(x) to solve these? I don't know where to start.
Last edited:
Physics news on Phys.org
  • #2
Hi again LizzieL! :smile:

Yes. Your expression for f'(x) is correct!

For problem 2 you'd simply substitute v=0.89 for x in f'(x).

For problem 3 you need to substitute w=0.5 for x in f(x) and in f'(x).
After that you need to find A and B such that f'(w) = A and f(w) = Aw + B.
  • #3
I got it! Thanks :approve:

Suggested for: Differentiation and tangent line