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Math Help - derivative problem and MVT - need explanation

  1. #1
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    Smile derivative problem and MVT - need explanation

    hey everyone i had 2 questions

    firstly
    If y = sqrt(x^2 + 1) , then the derivative of y^2 with respect to x^2 is

    i'm not sure where to start and its a mc question so i tried to first take the derivative of x^2 and with respect to dy/dx but i don't think that is right

    secondly
    the mean value theorem states that f(x) has to be continuous at [a,b], a and b being the end points, and f(x) has to be differentiable at (a,b), again a and b being the end points. if this is true, are the end points included in the mean value theorem

    thanks alot
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  2. #2
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    Re: derivative problem and MVT - need explanation

    Hello, ubhutto!

    \text{Given: }\:y \:= \sqrt{x^2 + 1}
    \text{Find the derivative of }y^2\text{ with respect to }x^2.

    Square both sides: . y^2 \:=\:x^2+1

    Let u = y^2,\;v = x^2

    Then we have: . u \:=\:v+1

    Hence: . \frac{du}{dv} \:=\:1

    Back-substitute: . \frac{d(y^2)}{d(x^2)} \:=\:1
    Thanks from ubhutto
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  3. #3
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    Re: derivative problem and MVT - need explanation

    With regards to your second question: yes they are included. The conditions ask for f to be continuous on [a,b]. This is does not mean f is continuous at a and b, (it does mean f should be defined at those points) so we must allow f to be non-differentiable at a and b, hence the open interval. For instance, the function:

    f(x) = \left\{  \begin{array}{l l} x^2 & \quad -1\leq x\leq 1\\0 & \quad \text{otherwise}\end{array} \right.

    is continuous on [-1,1] but discontinuous ( \to not differentiable) at -1 and 1. The conditions mean that we may apply the theorem on [-1,1] or any closed interval contained in [-1,1].
    Thanks from ubhutto
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