Differentiation and tangent line

  • Thread starter LizzieL
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  • #1
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Homework Statement


Function
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

So,
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:

Answers and Replies

  • #2
I like Serena
Homework Helper
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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
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I got it! Thanks :approve:
 

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