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Math Help - find the derivative

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by vance View Post
    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 k'(1) = 2.


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

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

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


    Use a similar approach for the rest...
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  3. #3
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    Quote Originally Posted by vance View Post
    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?
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