Hrmm derivative problems (concept?)

1. Oct 30, 2003

VikingStorm

I've been trying to do these concept-based questions, (but I think my concept isn't that sound).

"Suppose f'(2)=4, g'(2)=3, f(2)=-1 and g(2)=1. Find the derivative at 2 of each of the following functions
a. s(x)=f(x)+g(x)
b. p(x)=f(x)g(x)
c. q(x)=f(x)/g(x)"
I began doing this, without reading the find the derivative part. What order would I exactly solve it in? Or does it work straight in by plugging in the derivatives? (too simple, so must not be it)

"If f(x)=x, find f'(137)"
This is a pure concept question I'm sure...

"Explain what is wrong with the equation (x^2-1)/(x-1)=x+1, and why lim(x^2)/(x-1)=lim(x+1) both x->1"
The top factors out and supposedly cancels, though I'm not sure why I can't do that.

2. Oct 30, 2003

Soroban

Hello, VikingStorm!

"Suppose f'(2)=4, g'(2)=3, f(2)=-1 and g(2)=1. Find the derivative at 2 of each of the following functions
a. s(x) = f(x) + g(x)
b. p(x) = f(x)*g(x)
c. q(x) = f(x)/g(x)"

Yes , you're right ...
After finding the derivative, just plug in the given values.

(a) s'(x) = f '(x) + g'(x)
Hence: s'(2) = f'(2) + g'(2) = 4 + 3 = 7

(b) p'(x) = f(x)*g'(x) + g(x)*f '(x)
Hence: p'(2) = f(2)*g'(2) + g(2)*f '(2) = (-1)(3) + (1)(4) = 1

(c) q'(x) = [g(x)*f '(x) - f(x)*g'(x)]/[g(x)]^2
Hence: q'(2) = [(1)(4) - (-1)(3)][1^2] = 7

3. Nov 2, 2003

phoenixthoth

for the first one, note that f'(x)=1 for all x, so f'(137)=1. another way to look at is is that for y=x, y=x is a tangent line at all points. the slope of the tangent line is 1 everywhere, so since f'(x) is the slope of the tangent line at (x,f(x)), f'(137)=1.

for the second question, the main thing is what is meant by the equality sign. suppose A(x) and B(x) are two algebraic expressions defined for some set such as the set of real numbers. then we say that A(x)=B(x) if and only if A(x) equals B(x) for all real numbers x. such equations like A(x)=B(x) that are true "everywhere" are called identities.

(x^2-1)/(x-1)=x+1 is *not* an identity because the equation isn't always true: it fails when x=1.

if you let A(x)=(x^2-1)/(x-1) and B(x)=x+1, note that A(x)=B(x) for all real numbers except x=1. when you take the limit as x approaches 1, x is never allowed to actually equal 1, so
limA(x)=limB(x).