Calculus Small Changes, Big Ideas

Section 1.2 Basic differentiation techniques

• We ‘should believe’ that if \(y=mx+b\text{,}\) then

  \begin{align*} \dv{y}{x} \amp = m\\ \dd{y}\amp = m\dd{x} \end{align*}

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• If \(f\) changes by a small amount \(\dd{f}\) and \(g\) changes by a small amount \(\dd{g}\text{,}\) we should expect the linearity properties to hold:

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  \begin{align*} \dd(u+v) \amp = \dd{u}+\dd{v}\\ \dd(cu) \amp = c\dd{u} \end{align*}
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• Derivative of \(x^2\) can be done by visualizing a square of side length \(x\) and seeing how the area changes as \(x\) changes. You get two rectangles of area \(x\dd{x}\) and a small square of area \((\dd{x})^2\text{,}\) which is ‘negligible’, so the total change in area is approximately \(2x\dd{x}\text{,}\) and we imagine that if the change were infinitesimal, the change in area would be exactly \(2x\dd{x}\text{.}\)

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  We can temporarily justify this sense of being ‘negligible’ by looking at the ratios of changes:

  \begin{equation*} \dfrac{\Delta(x^2)}{\Delta x} = 2x + \Delta x \end{equation*}

  If we imagine that \(\Delta x\) becomes infinitesimal, the second term drops out, so we say that \(\DS\dv{(x^2)}{x} = 2x\text{.}\)

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  We’ll make this sense of ‘negligibility’ much more precise later when we introduce limits, but for now, we can just rely on intuition and the idea that if \(\dd{x}\) is very small, then \((\dd{x})^2\) is even smaller and can be ignored.

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• Same thing can be done with \(x^3\text{,}\) by visualizing a cube of side length \(x\) and seeing how the volume changes as \(x\) changes. You get three rectangular boxes of volume \(x^2\dd{x}\) and three rectangular boxes of volume \(x(\dd{x})^2\) and a small cube of volume \((\dd{x})^3\text{.}\) The higher powers of \(\dd{x}\) are ‘negligible’, so the total change in volume is approximately \(3x^2\dd{x}\text{,}\) and we imagine that if the change were infinitesimal, the change in volume would be exactly \(3x^2\dd{x}\text{.}\)

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• Power Rule (as a conjecture right now):

  \begin{equation*} \dd(x^{n})=nx^{n-1}\dd{x} \end{equation*}

  (Only works when \(n\) is a natural number so far)

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• Derivative of a polynomial term by term... can verify the derivative done last section! If \(L(v)=0.0034v^2 + 0.15v\text{,}\) then

  \begin{equation*} \dv{L}{v}=0.0068v+0.15\text{,} \end{equation*}

  so \(L'(50)=0.0068\cdot 50 + 0.15=0.49\text{,}\) which is the same as the slope of the tangent line we found last section. But it’s much more direct and efficient.

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