1. ## find the derivative

I'm little confused with this excersise.....

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((1/4)x) when x=4

in this excersise , the chain rule should be applied, but I don't know how. Can I get some help?

2. Originally Posted by vance
I'm little confused with this excersise.....

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((1/4)x) when x=4

in this excersise , the chain rule should be applied, but I don't know how. Can I get some help?
You are told that $\displaystyle k'(1) = 2$.

Now use the chain rule on the other functions...

$\displaystyle \frac{d}{dx}[k(2x)] = 2k'(2x)$

Now let $\displaystyle x = \frac{1}{2}$.

Use a similar approach for the rest...

3. Originally Posted by vance
I'm little confused with this excersise.....

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

a k(2x) when x = 1/2?
Let u= 2x. Then [tex]\frac{dk(2x)}{dx}= \frac{dk(u)}{du}\frac{du}{dx} and, of course, when x= 1/2, u= 1.

b k(x+1) when x=0
Let u= x+1. When x= 0, u= 1.

c k((1/4)x) when x=4
Let u= (1/4)x. When x= 4, u= 1.

in this excersise , the chain rule should be applied, but I don't know how. Can I get some help?