Differentiation and tangent line

[LizzieL]

Homework Statement

Function
f(x) = K^xL

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) = (K^u)'(u) * u'

(K^u)'(u) = (e^{u ln K})' = ln K * e^{u ln K} = ln K * K^u
u' = Lx^L-1 = -1.39x^-2.39

So,
f'(x) = ln 1.78(1.78^x-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.

 

[I like Serena]

Hi again LizzieL!


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.

 

[LizzieL]

I got it! Thanks