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Math Help - Differentiation

  1. #1
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    Differentiation

    Hi
    I am working through some introductory calculus. so I have a question (see attachment Q 56.)

    . I got 16x from y=8x^2 as the gradient function. What I want to know is, am I supposed to determine the x coordinates from these two functions using the gradient function?

    I just graphed the two functions and found their intercepts for the coordinates thus found the gradients for the two points this way.

    I'd like help with Q 58 also please.
    Attached Thumbnails Attached Thumbnails Differentiation-h38fyddc8ar2d1svht2tb6y2-1-.jpg  
    Last edited by Tren301; February 2nd 2014 at 12:03 AM.
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  2. #2
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    Re: Differentiation

    Solve the quadratic equation 8x^2=8x-16 to find the x coordinates of the intersection points.
    you will find -1 and 8 . Substitute these values of x into the 16x to get the two values of the gradient.......
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  3. #3
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    Re: Differentiation

    Firstly we know that the derivative gives us the gradient of the tangent to the curve at a particular point.
    Let us consider your question, Q No 58.
    equation of the curve is y = a x^4 Differentiating w r t x we get
    y' = 4ax^3. This gives us slope of tangent to the curve. It is given that the gradient of the tangent at ( 3,b) is 2.
    Gradient at ( 3,b) = 4 * a * 3 ^ 3 = 27*4 a = 2 given That gives 2 = 1/54
    Also point ( 3, b ) lies on the curve and so must satisfy it
    Thus we have from y = a * x^ 4
    b = a * 3 ^ 4= 81 a OR b = 1/ 54 * 81 = 3/2. Hence we get a = 1/ 54 and b = 3/2
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