Answers to exercises
Exercise 1
Let \(f(x) = x^4\). We first compute
\[ f(-1 + \Delta x) = (-1 + \Delta x)^4 = 1 - 4 (\Delta x) + 6 (\Delta x)^2 - 4 (\Delta x)^3 + (\Delta x)^4 \]and \(f(-1) = 1\). The gradient at \(x=-1\) is then given by
\begin{align*} \lim_{\Delta x \to 0} \dfrac{f(-1+\Delta x) - f(-1)}{\Delta x} &= \lim_{\Delta x \to 0} \dfrac{-4 (\Delta x) + 6 (\Delta x)^2 - 4 (\Delta x)^3 + (\Delta x)^4}{\Delta x} \\\\ &= \lim_{\Delta x \to 0} \bigl( -4 + 6 (\Delta x) - 4 (\Delta x)^2 + (\Delta x)^3 \bigr) = -4. \end{align*}The tangent line at \((-1,1)\) has gradient \(-4\), and hence has equation \(y - 1 = -4(x+1)\) or, equivalently, \(y = -4x - 3\).
Exercise 2
Let \(f(x) = x^3\). We compute
\begin{align*} f'(x) &= \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \dfrac{(x+\Delta x)^3 - x^3}{\Delta x} \\\\ &= \lim_{\Delta x \to 0} \dfrac{3 x^2 (\Delta x) + 3 x (\Delta x)^2 + (\Delta x)^3}{\Delta x} = \lim_{\Delta x \to 0} \bigl( 3x^2 + 3x (\Delta x) + (\Delta x)^2 \bigr) = 3x^2. \end{align*}Exercise 3
Let \(f(x) = c\). Then, for any \(x\) and \(\Delta x\), we have \(f(x+\Delta x) - f(x) = c-c = 0\). Hence, the derivative is
\[ f'(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} 0 = 0. \]Exercise 4
Let \(f(x) = ax + b\). Then
\begin{align*} f'(x) &= \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \dfrac{a(x+\Delta x) + b - ax - b}{\Delta x} \\\\ &= \lim_{\Delta x \to 0} \dfrac{a (\Delta x)}{\Delta x} = \lim_{\Delta x \to 0} a = a. \end{align*}Exercise 5
Let \(f(x) = \dfrac{1}{x}\). We first compute the quotient
\begin{align*} \dfrac{f(x+\Delta x)-f(x)}{\Delta x} &= \dfrac{1}{\Delta x} \Bigl( \dfrac{1}{x+\Delta x} - \dfrac{1}{x} \Bigr) = \dfrac{1}{\Delta x} \cdot \dfrac{x - (x+\Delta x)}{x (x+\Delta x)} \\\\ &= \dfrac{-\Delta x}{(\Delta x) x (x+\Delta x)} = -\dfrac{1}{x(x+\Delta x)}. \end{align*}Hence, the derivative is
\[ f'(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} - \dfrac{1}{x(x+\Delta x)} = - \dfrac{1}{x^2}. \]Thus the derivative of \(f(x) = \dfrac{1}{x}\) is \(f'(x) = - \dfrac{1}{x^2}\).
Exercise 6
The derivative of \(f(x) - g(x)\) is given by
\begin{align*} & \lim_{\Delta x \to 0} \dfrac{\bigl( f(x+\Delta x) - g(x+\Delta x) \bigr) - \bigl( f(x) - g(x) \bigr)}{\Delta x} \\\\ ={}& \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} - \lim_{\Delta x \to 0} \dfrac{g(x+\Delta x) - g(x)}{\Delta x} \\\\ ={}& f'(x) - g'(x). \end{align*}Hence, the derivative of \(f(x) - g(x)\) is \(f'(x) - g'(x)\).
Exercise 7
Rewriting \(f(x)\) as \(x^3 + x^{-\frac{3}{2}}\), we obtain \(f'(x) = 3x^2 - \dfrac{3}{2} x^{-\frac{5}{2}}\).
Exercise 8
We assume that the derivative of \(x\) is 1. We now prove that, if the derivative of \(x^n\) is \(nx^{n-1}\), then the derivative of \(x^{n+1}\) is \((n+1)x^n\). To do this we use the product rule:
\begin{align*} \dfrac{d}{dx} \bigl( x^{n+1} \bigr) &= \dfrac{d}{dx} \bigl( x^n \cdot x \bigr) = x^n \dfrac{d}{dx} \bigl( x \bigr) + x \dfrac{d}{dx} \bigl( x^n \bigr) \\ &= x^n \cdot 1 + x \cdot nx^{n-1} = (n+1) x^n. \end{align*}It follows by induction that \(\dfrac{d}{dx}(x^n) = nx^{n-1}\), for all positive integers \(n\).
Exercise 9
Using the product rule, we have
\begin{align*} \dfrac{d}{dx} f(x)^2 &= \dfrac{d}{dx} \bigl( f(x) \cdot f(x) \bigr) = f(x) \cdot \dfrac{d}{dx} \bigl( f(x) \bigr) + f(x) \cdot \dfrac{d}{dx} \bigl( f(x) \bigr) \\\\ &= f(x)\,f'(x) + f(x)\,f'(x) = 2\,f(x)\,f'(x). \end{align*}Exercise 10
We first use the product rule on the product of \(f(x)\,g(x)\) and \(h(x)\):
\[ \dfrac{d}{dx} \bigl[ f(x)\,g(x)\,h(x) \bigr] = \dfrac{d}{dx} \bigl[ f(x)\,g(x) \bigr]\,h(x) + f(x)\,g(x)\,\dfrac{d}{dx} \bigl[ h(x) \bigr]. \]Then we use the product rule on \(f(x)\,g(x)\):
\begin{align*} \dfrac{d}{dx} \bigl[ f(x)\,g(x)\,h(x) \bigr] &= \bigl( f'(x)\,g(x) + f(x)\,g'(x) \bigr)\,h(x) + f(x)\,g(x)\,h'(x) \\\\ &= f'(x)\,g(x)\,h(x) + f(x)\,g'(x)\,h(x) + f(x)\,g(x)\,h'(x). \end{align*}For a general product \(f_1(x)\,f_2(x) \dotsm f_n(x)\), the derivative is a sum of \(n\) terms, with \(f'_i(x)\) occurring in the \(i\)th term:
\begin{align*} \dfrac{d}{dx} \bigl[ & f_1(x)\,f_2(x) \dotsm f_n(x) \bigr] \\\\ &= f'_1(x)\,f_2(x) \dotsm f_n(x) + f_1(x)\,f'_2(x)\,f_3(x) \dotsm f_n(x) + \dots + f_1(x) \dotsm f_{n-1}(x)\,f'_n(x). \end{align*}Exercise 11
Let \(f(x) = (x^2 + 7)^{100} = g(h(x))\), where \(g(x) = x^{100}\) and \(h(x) = x^2 + 7\). Then \(g'(x) = 100x^{99}\) and \(h'(x) = 2x\), so by the chain rule
\[ f'(x) = g'(h(x))\,h'(x) = 100 (x^2 + 7)^{99} \cdot 2x = 200 x (x^2 + 7)^{99}. \]Exercise 12
We can write \(f(x)^2\) as \(h(f(x))\) where \(h(x) = x^2\). The chain rule then gives
\[ \dfrac{d}{dx} h(f(x)) = h'(f(x))\,f'(x) = 2\,f(x)\,f'(x). \]Exercise 13
We first think of \(f(g(h(x)))\) as the composition of \(f(x)\) and \(g(h(x))\), so the chain rule gives
\[ \dfrac{d}{dx} \bigl[ f(g(h(x))) \bigr] = f'(g(h(x)))\,\dfrac{d}{dx} g(h(x)). \]Then using the chain rule again gives
\[ \dfrac{d}{dx} \bigl[ f(g(h(x))) \bigr] = f'(g(h(x)))\,g'(h(x))\,h'(x). \]In general, for the composition of \(n\) functions \(f_1 \circ f_2 \circ \dots \circ f_n\), the derivative is a product of \(n\) factors, and the \(i\)th factor is \(f'_i ( f_{i+1} ( \dotsb ( f_n (x) ) \dotsb ))\).
Exercise 14
Let \(g(x) = x^n\) and \(h(x) = \dfrac{1}{x}\), so that \(\dfrac{1}{x^n} = h(g(x))\). Then \(g'(x) = nx^{n-1}\) and \(h'(x) = - \dfrac{1}{x^2}\). By the chain rule,
\begin{align*} \dfrac{d}{dx} \Biggl( \dfrac{1}{x^n} \Biggr) &= \dfrac{d}{dx} h(g(x)) = h'(g(x))\,g'(x) \\ &= -\dfrac{1}{(g(x))^2}\,g'(x) = -\dfrac{1}{x^{2n}}\,n x^{n-1} = - nx^{-n-1}. \end{align*}Thus, the derivative of \(x^{-n}\) is \(-nx^{-n-1}\).
Exercise 15
Let \(h(x) = \dfrac{1}{x}\). Then \(h'(x) = - \dfrac{1}{x^2}\) and \(\dfrac{1}{g(x)} = h(g(x))\). By the chain rule,
\[ \dfrac{d}{dx} \Biggl( \dfrac{1}{g(x)} \Biggr) = \dfrac{d}{dx} h(g(x)) = h'(g(x))\,g'(x) = -\dfrac{1}{(g(x))^2}\,g'(x) = -\dfrac{g'(x)}{(g(x))^2}. \]Exercise 16
Since the derivative of \(\dfrac{1}{g(x)}\) is \(-\dfrac{g'(x)}{(g(x))^2}\), the product rule gives
\begin{align*} \dfrac{d}{dx} \Biggl( \dfrac{f(x)}{g(x)} \Biggr) &= \dfrac{1}{g(x)} \, \dfrac{d}{dx} \bigl(f(x)\bigr) + f(x) \, \dfrac{d}{dx} \Biggl( \dfrac{1}{g(x)} \Biggr) \\\\ &= \dfrac{f'(x)}{g(x)} + f(x) \Bigl( -\dfrac{g'(x)}{(g(x))^2} \Bigr) = \dfrac{g(x)\,f'(x) - f(x)\,g'(x)}{(g(x))^2}. \end{align*}This is the quotient rule.
Exercise 17
Let \(y = \sqrt{9 - x^2}\). We differentiate:
\[ \dfrac{dy}{dx} = \dfrac{1}{2} (9-x^2)^{-\dfrac{1}{2}} \cdot (-2x) = \dfrac{-x}{\sqrt{9-x^2}}. \]So, at \(x = \dfrac{3 \sqrt{2}}{2}\), we have \(y = \dfrac{3 \sqrt{2}}{2}\) and \(\dfrac{dy}{dx} = \dfrac{-3 \sqrt{2}}{2} \dfrac{2}{3 \sqrt{2}} = -1\). The tangent line has gradient \(-1\) and passes through the point \((\dfrac{3 \sqrt{2}}{2}, \dfrac{3\sqrt{2}}{2})\), and hence has equation \(y=-x+ 3\sqrt{2}\).
Exercise 18
From \(f(x) = (x^2+7)^{100}\), we have \(f'(x)=200x(x^2 + 7)^{99}\). Using the product and chain rules, we obtain
\begin{align*} f''(x) &= 200x\,\dfrac{d}{dx}\bigl[(x^2+7)^{99}\bigr] + (x^2+7)^{99}\,\dfrac{d}{dx}\bigl[200x\bigr] \\\\ &= 200 x \cdot 99 (x^2+7)^{98} \cdot 2x + (x^2 + 7)^{99} \cdot 200 \\\\ &= 200 (x^2 + 7)^{98} (199x^2 + 7). \end{align*}Exercise 19
We compute the derivatives of \(x(t) = 1 - 7t + (t-5)^4\) with respect to \(t\):
\begin{align*} x'(t) &= -7 + 4(t-5)^3 \\\\ x''(t) &= 12(t-5)^2. \end{align*}Since squares are non-negative, we have \(x''(t) \geq 0\) for all \(t\). That is, the acceleration is always non-negative.
Exercise 20
Let \(y = x^{\frac{1}{n}}\), where \(n\) is a positive integer. We wish to find \(\dfrac{dy}{dx}\). We have \(x = y^n\), and so \(\dfrac{dx}{dy} = ny^{n-1}\). Thus
\[ \dfrac{dy}{dx} = \dfrac{1}{\dfrac{dx}{dy}} = \dfrac{1}{ny^{n-1}} = \dfrac{1}{n} y^{1-n}, \]and substituting \(y = x^{\frac{1}{n}}\) gives
\[ \dfrac{dy}{dx} = \dfrac{1}{n} x^{\frac{1-n}{n}} = \dfrac{1}{n} x^{\frac{1}{n} - 1}, \]as expected.
Exercise 21
Let \(y = x^{\frac{p}{q}}\), where \(p,q\) are integers with \(q > 0\). We wish to find \(\dfrac{dy}{dx}\). We let \(u = x^{\frac{1}{q}}\). Then \(y = u^p\) and, by the previous exercise, \(\dfrac{du}{dx} = \dfrac{1}{q} x^{\frac{1}{q} - 1}\). The chain rule then gives
\[ \dfrac{dy}{dx} = \dfrac{dy}{du} \, \dfrac{du}{dx} = pu^{p-1} \cdot \dfrac{1}{q} x^{\frac{1}{q} - 1} = \dfrac{p}{q} u^{p-1} x^{\frac{1}{q} - 1}. \]Substituting \(u = x^{\frac{1}{q}}\) gives
\[ \dfrac{dy}{dx} = \dfrac{p}{q} x^{\frac{p-1}{q}} x^{\frac{1}{q} - 1} = \dfrac{p}{q} x^{\frac{p}{q} - 1}. \]Exercise 22
If the circle is centred at the origin, then the radius of the circle from \((0,0)\) to \((x,y)\) has gradient \(\dfrac{y}{x}\). The tangent to the circle is perpendicular to the radius, and hence its gradient is the negative reciprocal of \(\dfrac{y}{x}\), that is, the gradient is \(-\dfrac{x}{y}\).
Exercise 23
- From \(y^2 = x^2 - 5\), we have \(y = \pm \sqrt{x^2-5}\). As we want to include the point \((3,-2)\), we take the negative square root and consider \(y = - \sqrt{x^2 - 5}\). Then \[ \dfrac{dy}{dx} = - \dfrac{1}{2} (x^2 - 5)^{-\dfrac{1}{2}} \cdot 2x = \dfrac{-x}{\sqrt{x^2-5}}. \] At \(x=3\), we have \(\dfrac{dy}{dx} = \dfrac{-3}{\sqrt{4}} = -\dfrac{3}{2}\).
- Implicit differentiation of \(x^2 - y^2 = 5\) gives \(2x - 2y\,\dfrac{dy}{dx} = 0\), and so \(\dfrac{dy}{dx} = \dfrac{x}{y}\). Hence, at the point \((3,-2)\), we have \(\dfrac{dy}{dx} = -\dfrac{3}{2}\).
Exercise 24
The \(k\)th derivative of \(x^n\) is \(n (n-1) \dotsb (n-k+1) x^{n-k}\), and hence the \(n\)th derivative is the constant \(n!\).
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