# Need Help Understanding

1. Oct 28, 2005

### Tom McCurdy

If the derivative of y=k(x) equals 2 when x=1 what is the derivative of

a. k(2x) when x = 1/2
b. k(x+1) when x=0
c. k(x/4) when x =4

I had two different ideas both seemed wrong
the first idea was to implore some type of chain rule so
a. 2*2=4
b. 2*1=2
c 2*1/4= 1/2

by saying the outside function would be 2 and then taking the derivative of the inside function, but you can't really do this so it seems wrong

the second idea seems even more wrong to simpliy before the derivative so that everything would come to be k(x) making a,b, and c =2

if someone could help me that would be great

thx
tom

2. Oct 28, 2005

### kreil

this problem seems as straightforward as your second idea i think..

i mean if you evaluate something at f(1) or f(2/2) its the same thing..

but then again, why would they ever ask that if this is the case?

Josh

3. Oct 28, 2005

### HallsofIvy

Staff Emeritus
"employ", not "implore" (unless you are begging the chain rule to work!)
The "outside function" is not 2- the inside function is 2x.
The derivative of k(2x) is k'(2x)(2) where "k'(2x)" means just the derivative of k(x) evaluated at 2x. Since y'(1)= 2, y'(2(1/2))= y'(1)= 2 and so
the derivative of k(2x) at x= 1/2 is 2(2)= 4.

If by "simplify" you mean "substitute", yes, that's a great idea. You should eventually learn to apply the chain rule quickly without having to write down the substitution but it's good practice to write it out while you are still learning.
In (a), let u= 2x. Then dk/dx= (dk/du)(du/dx). du/dx= 2 and u= 1 when x= 1/2 so the derivative is (2)(2)= 4.

In (b), let u= x+ 1. Now du/dx= 1 and u= 1 when x= 0 so the derivative is
(2)(1)= 2.

In (c), let u= x/4. Now du/dx= 1/4 and u= 1 when x= 4 so the derivative is
(2)(1/4)= 1/2.