Calculus Small Changes, Big Ideas

Section 1.4 Implicit differentiation

• Use Bitcoin elliptic curve example?

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• Use differentials: take differential of both sides, then divide both sides by \(\dd{x}\text{.}\)

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  Example:

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  \begin{align*} x^2 + y^2 \amp = 1 \\ \dd(x^2 + y^2) \amp = \dd(1) \\ 2x\dd{x} + 2y\dd{y} \amp = 0 \\ 2x\dv{x}{x} + 2y\dv{y}{x} \amp = 0 \\ 2x + 2y\dv{y}{x} \amp = 0 \\ 2y\dv{y}{x} \amp = -2x \\ \dv{y}{x} \amp = -\frac{x}{y} \end{align*}
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• You can shortcut the differentials by taking derivatives and remembering to multiply by \(\DS\dv{y}{x}\) whenever you take a derivative of something involving \(y\text{.}\)

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• Talk about tangent lines. DON’T include normal lines; save that until the Analytic Geometry chapter in Part II.

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• Differentiating \(\dfrac{1}{x}\text{:}\)

  \begin{align*} y \amp = \frac{1}{x} \\ xy \amp = 1 \\ \dd(xy) \amp = \dd(1) \\ x\dd{y} + y\dd{x} \amp = 0 \\ x\dd{y} \amp = -y\dd{x} \\ \dd{y} \amp = -\frac{y}{x}\dd{x} \\ \dd{y} \amp = -\frac{1}{x^2}\dd{x} \\ \dd(x^{-1}) \amp = -1x^{-2}\dd{x} \end{align*}

  Can leave differentiating \(\dfrac{1}{x^n}\) as an exercise, but state the result that the Power Rule now works for all negative exponents.

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• Develop Quotient Rule:

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  \begin{align*} h \amp = \frac fg\\ gh \amp = f \\ \dd(gh) \amp = \dd{f} \\ g\dd{h} + h\dd{g} \amp = \dd{f} \\ g\dd{h} \amp = \dd{f} - h\dd{g} \\ \dd{h} \amp = \frac{\dd{f} - h\dd{g}}{g} \\ \dd{h} \amp = \frac{\dd{f} - \frac{f}{g}\dd{g}}{g}\cdot \frac{g}{g} \\ \dd(\frac fg) \amp = \frac{g\dd{f} - f\dd{g}}{g^2} \end{align*}
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• Differentiating \(\sqrt{x}\text{:}\)

  \begin{align*} y \amp = \sqrt{x} \\ y^2 \amp = x \\ \dd(y^2) \amp = \dd{x} \\ 2y\dd{y} \amp = \dd{x} \\ \dd{y} \amp = \frac{1}{2\sqrt{x}}\dd{x} \\ \dd(x^{1/2}) \amp = \frac{1}{2}x^{-1/2}\dd{x} \end{align*}

  Can leave differentiating \(\sqrt[b]{x^a}\) as an exercise, but state the result that the Power Rule now works for all fractional exponents.

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