CLP-1 Differential Calculus

Compute the derivative, \(f'(a)\text{,}\) of the function \(f(x)=\sqrt{x}\) at the point \(x=a\) for any \(a \gt 0\text{.}\)

• So again we start with the definition of derivative and go from there:
  \begin{align*} f'(a) &=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} =\lim_{x\rightarrow a}\frac{\sqrt{x}-\sqrt{a}}{x-a} \end{align*}
• As \(x\) tends to \(a\text{,}\) the numerator and denominator both tend to zero. But \(\tfrac{0}{0}\) is not defined. So to get a well defined limit we need to exhibit a cancellation between the numerator and denominator — just as we saw in Examples 1.4.12 and 1.4.17. Now there are two equivalent ways to proceed from here, both based on a similar “trick”.
• For the first, review Example 1.4.17, which concerned taking a limit involving square-roots, and recall that we used “multiplication by the conjugate” there:
  \begin{align*} \amp\frac{\sqrt{x}-\sqrt{a}}{x-a}\\ &= \frac{\sqrt{x}-\sqrt{a}}{x-a} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}} && \Big(\text{multiplication by $1=\frac{\text{conjugate}}{\text{conjugate}}$}\Big)\\ &=\frac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})} {(x-a)(\sqrt{x}+\sqrt{a})}\\ &= \frac{x-a}{(x-a)(\sqrt{x}+\sqrt{a})} && \big(\text{since $(A-B)(A+B) = A^2-B^2$)}\,\big)\\ &= \frac{1}{\sqrt{x}+\sqrt{a}} \end{align*}
• Alternatively, we can arrive at \(\frac{\sqrt{x}-\sqrt{a}}{x-a}=\frac{1}{\sqrt{x}+\sqrt{a}}\) by using almost the same trick to factor the denominator. Just set \(A=\sqrt{x}\) and \(B=\sqrt{a}\) in \(A^2 - B^2 = (A-B)(A+B) \) to get
  \begin{align*} x - a &= (\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a}) \end{align*}
  and then substitute this little fact into our expression
  \begin{align*} \frac{\sqrt{x}-\sqrt{a}}{x-a} &=\frac{\sqrt{x}-\sqrt{a}}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})} & \text{(now cancel common factors)}\\ &=\frac{1}{(\sqrt{x}+\sqrt{a})} \end{align*}
• Once we know that \(\frac{\sqrt{x}-\sqrt{a}}{x-a}=\frac{1}{\sqrt{x}+\sqrt{a}}\text{,}\) we can take the limit we need:
  \begin{align*} f'(a) &=\lim_{x\rightarrow a}\frac{\sqrt{x}-\sqrt{a}}{x-a}\\ & =\lim_{x\rightarrow a}\frac{1}{\sqrt{x}+\sqrt{a}}\\ & =\frac{1}{2\sqrt{a}} \end{align*}

• We should think about the domain of \(f'\) here — that is, for which values of \(a\) is \(f'(a)\) defined? The original function \(f(x)\) was defined for all \(x \geq 0\text{,}\) however the derivative \(f'(a)=\frac{1}{2\sqrt{a}}\) is undefined at \(a = 0\text{.}\)

  If we draw a careful picture of \(\sqrt{x}\) around \(x=0\) we can see why this has to be the case. The figure below shows three different tangent lines to the graph of \(y=f(x)=\sqrt{x}\text{.}\) As the point of tangency moves closer and closer to the origin, the tangent line gets steeper and steeper. The slope of the tangent line at \(\big(a,\sqrt{a}\big)\) blows up as \(a\to 0\text{.}\)

  

Compute the derivative, \(f'(a)\text{,}\) of the function \(f(x)=|x|\) at the point \(x=a\text{.}\)

• We should start this example by recalling the definition of \(|x|\) (we saw this back in Example 1.5.6):

  \begin{align*} |x| &= \begin{cases} -x & \text{ if } x \lt 0\\ 0 & \text{ if } x=0\\ x & \text{ if }x \gt 0. \end{cases} \end{align*}

  It is definitely not just “chop off the minus sign”.

  
• This breaks our computation of the derivative into 3 cases depending on whether \(x\) is positive, negative or zero.
• Assume \(x \gt 0\text{.}\) Then
  \begin{align*} \diff{f}{x} &= \lim_{h\to0} \frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0} \frac{|x+h|-|x|}{h}\\ \\ \end{align*}

  Since \(x \gt 0\) and we are interested in the behaviour of this function as \(h \to 0\) we can assume \(h\) is much smaller than \(x\text{.}\) This means \(x+h \gt 0\) and so \(|x+h|=x+h\text{.}\)

  \begin{align*} &= \lim_{h\to0} \frac{x+h-x}{h}\\ &= \lim_{h\to0} \frac{h}{h} = 1 &\text{as expected} \end{align*}
• Assume \(x \lt 0\text{.}\) Then
  \begin{align*} \diff{f}{x} &= \lim_{h\to0} \frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0} \frac{|x+h|-|x|}{h}\\ \\ \end{align*}

  Since \(x \lt 0\) and we are interested in the behaviour of this function as \(h \to 0\) we can assume \(h\) is much smaller than \(x\text{.}\) This means \(x+h \lt 0\) and so \(|x+h|=-(x+h)\text{.}\)

  \begin{align*} &= \lim_{h\to0} \frac{-(x+h)-(-x)}{h}\\ &= \lim_{h\to0} \frac{-h}{h} = -1 \end{align*}
• When \(x=0\) we have
  \begin{align*} f'(0) &= \lim_{h\to0} \frac{f(0+h)-f(0)}{h}\\ &= \lim_{h\to0} \frac{|0+h|-|0|}{h}\\ &= \lim_{h\to0} \frac{|h|}{h} \end{align*}
  To proceed we need to know if \(h \gt 0\) or \(h \lt 0\text{,}\) so we must use one-sided limits. The limit from above is:
  \begin{align*} \lim_{h \to 0^+} \frac{|h|}{h} &= \lim_{h \to 0^+} \frac{h}{h} &\text{since } h \gt 0, |h|=h\\ &= 1\\ \end{align*}

  Whereas, the limit from below is:

  \begin{align*} \lim_{h \to 0^-} \frac{|h|}{h} &= \lim_{h \to 0^-} \frac{-h}{h} &\text{since } h \lt 0, |h|= -h\\ &= -1 \end{align*}
  Since the one-sided limits differ, the limit as \(h\to 0\) does not exist. And thus the derivative does not exist as \(x=0\text{.}\)

In summary: