# Thread: some explanations

1. ## some explanations

Hello,

I am sorry that I cannot put a more descriptive title but the fact is that the info in usual titles is what I am looking for... sort of.

Any who, I have a certain question type to answer and I am in fact quite lost. I will give an example but I am NOT looking for an answer. I am simply looking for a guideline.

Code:
Find h'(2) given that f(2) = -5, f '(2) = 2, g(2) = 6, and g'(2) = 5.
(a) h(x) = 2f(x) - 2g(x)
h'(2) = ???
Cheers and thanks!

An

2. ## easy so check it out

f(2)=-5
f'(2)=2
f'(2)=limit[f(2+t)-f(2)]/t
t->0

g(2)=6
g'(2)=5
g'(2)=limit[g(2+t)-g(2)]/t
t->0

h(x)=2f(X)-2g(x)
h'(x)=limit[h(x+t)-h(x)]/t
t->0
but
h(2+t)=2f(2+t)-2g(2+t)
and h(2)=2f(2)-2g(2)
therfor
h'(x)=limit[2f(2+t)-2g(x+t)-(2f(x)-2g(x))]/t
t->0
or
h'(2)= 2xlimit[f(2+t)-f(2)]/t-2xlimit[g(2+t)-g(2)]/t
t->0 t->0
putting the values we get
h'(2)= 2[2-5]
=-6
note : due to functional equation we don't need the value of f(2) and g(2)

3. Originally Posted by Anhim
Hello,

I am sorry that I cannot put a more descriptive title but the fact is that the info in usual titles is what I am looking for... sort of.

Any who, I have a certain question type to answer and I am in fact quite lost. I will give an example but I am NOT looking for an answer. I am simply looking for a guideline.

Code:
Find h'(2) given that f(2) = -5, f '(2) = 2, g(2) = 6, and g'(2) = 5.
(a) h(x) = 2f(x) - 2g(x)
h'(2) = ???
Cheers and thanks!

An
From the linear property of the derivative (which you ought to know):

h'(x) = 2 f'(x) - 2 g'(x) therefore h'(2) = 2 f'(2) - 2 g'(2). Now substitute the given values.

4. oh my dear... How did I not see that before?!

I knew it would be something easy as such but until you know for sure...

Well thanks!

An