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Thread: How do I find the derivative of this equation?

  1. #16
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    Re: How do I find the derivative of this equation?

    Thank you, that is a stellar explanation.
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  2. #17
    MHF Contributor MarkFL's Avatar
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    Re: How do I find the derivative of this equation?

    Consider the following general quadratic:

    $\displaystyle y=ax^2+bx+c$

    Now, as I stated, the axis of symmetry, which will be the value of $\displaystyle x$ at the vertex, is"

    $\displaystyle x=-\frac{b}{2a}$

    This is found by equating the discriminant to zero, or equivalently, finding the mean of the roots, as given by the quadratic formula.

    Using differential calculus, we may equate the first derivative to zero, to find critical values of $\displaystyle x$, that is, for which values of $\displaystyle x$ are there local extrema, or turning points.

    $\displaystyle \frac{dy}{dx}=2ax+b=0\implies x=-\frac{b}{2a}$

    Unless you understand the definition of the derivative, and why equating it to zero leads to the potential critical values/turning points and how to use derivative-based tests to determine the nature of the extrema, you really are just applying a tool you're simply not ready to use. When we use formulas to solve problems, we really should understand why they work, and even be able to derive them.
    Thanks from topsquark
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