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?
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.
Let u= x+1. When x= 0, u= 1.b k(x+1) when x=0
Let u= (1/4)x. When x= 4, u= 1.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?