##### 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
1. 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 $$\dfrac{dy}{dx} = \dfrac{-3}{\sqrt{4}} = -\dfrac{3}{2}$$.
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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