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Wize AP Calculus (AB) Textbook > Differentiation 1: Basics & Fundamental Properties

The Limit Definition of a Derivative

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The Limit Definition of the Derivative

The derivative is one of the most important and fundamental concepts in calculus. The process of differentiation tells us the rate of change of a function.


The Derivative (At a Point)

The derivative of a function f(x)f(x) at a point x=a,x=a, denoted f′(a)f'(a), is
f′(a)=lim⁡h→0f(a+h)−f(a)h\boxed{f'(a)=\displaystyle\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}}

When this limit exists, we say that the function is differentiable at x=a.x=a.

Note: The derivative f′(a)f'(a) is also denoted dfdx∣x=a\frac{df}{dx}\Bigg|_{x=a}


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Example: Definition of a Derivative


Using the definition of a derivative, find the derivative of f(x)=x.f(x)=\sqrt{x}.



f′(x)=lim⁡h→0x+h−xh⋅x+h+xx+h+x=lim⁡h→0x+h−xh(x+h+x)=lim⁡h→01x+h+x=12x\begin{array}{ll} f'(x) & =\displaystyle \lim_{h \to 0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\ & =\displaystyle \lim_{h \to 0}\dfrac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\\ & =\displaystyle \lim_{h \to 0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}}\\ & =\displaystyle \bf \dfrac{1}{2\sqrt{x}} \end{array}



Using the definition of the derivative, find the derivative of f(x)=(x+2)2+xf(x)=(x+2)^2+x




When is Function Not Differentiable?

Here are the 4 cases where a function is not differentiable (i.e. the limit lim⁡h→0f(a+h)−f(a)h\displaystyle\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h} does not exist)
  1. If a function is discontinuous at a point, then it is not differentiable at that point:
  2. Jump discontinuity: the function is not differentiable at the jump

  1. Hole (removable discontinuity): the function is not differentiable at the hole

  1. Vertical asymptote: the function is not differentiable at the x value where the vertical asymptote is
  1. A function is not differentiable at a corner or cusp: For example, the function f(x)=∣x∣f\left(x\right)=\left|x\right| is not differentiable at x=0x=0

  1. A function is not differentiable where there is a vertical tangent line: For example, the function f(x)=x13f\left(x\right)=x^{\frac{1}{3}} or x3\sqrt[3]{x} is not differentiable at x=0x=0

  1. A function is not differentiable if it oscillates back and forth as it approaches a certain point: For example, the function f(x)=sin⁡(1x)f(x)=\sin\left(\dfrac{1}{x}\right) is not differentiable at x=0x=0 because the ff value oscillates back and forth between 1 and -1.


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Checking Differentiability at a Point

A common differentiability question you might see on the exam is that they'll give you a piecewise function f(x)={g(x), x<ah(x), x≥af(x)=\begin{cases} g(x)&,~x<a\\ h(x)&,~x\ge a \end{cases} and ask if the function is differentiable at the connection point x=ax=a. For this function to be differentiable, we need to check that lim⁡h→0−f(a+h)−f(a)h=lim⁡h→0+f(a+h)−f(a)h\displaystyle \lim_{h\to 0^-}\dfrac{f(a+h)-f(a)}{h}=\displaystyle \lim_{h\to 0^+}\dfrac{f(a+h)-f(a)}{h}. However, this can be very time-consuming and hard to do. Instead, we can usually check the two following things:
  1. Check that lim⁡x→a−f(x)=lim⁡x→a+f(x)=f(a)\displaystyle \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a). Most of the time you can just check if g(a)=h(a)g(a)=h(a) (this is a commonly used trick that only works MOST of the time)
  2. Check that g′(a)=h′(a)g'(a)=h'(a). We can use differentiation rules to find g′(a)g'(a) and h′(a)h'(a) instead of having to evaluate a difficult limit. (See the next few sections to learn about these differentiation rules)