Exercises 2

This page allows you to practice some exercises on Differentiation. If you notice that you have difficulties, we advise you to go over the corresponding sections in the book.

These questions correspond to the following sections: - Section 6.5 - A Brief Introduction to Limits - Section 6.6 - Simple Rules for Differentiation - Section 6.7 - Sums, Products, and Quotients

Question 6

Determine the following limits by using rules for limits. (a) $\lim_{x \to 0} (3 + 2x^{2})$

(b) $\lim_{t \to 8} (5t + t^2 - \frac{1}{8}t^{3})$

(c) $\lim_{z \to -2} \frac{1/z + 2}{z}$

(d) $\lim_{y \to 0} \frac{(y + 1)^{5} - y^{5}}{y+1}$

(e) $\lim_{\lambda \to 0} \frac{1}{\lambda} (3^{\lambda} - 2^{\lambda})$

(f) $\lim_{x \to 1} \frac{x^2 + 7x - 8}{x - 1}$

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(a) $3$.

(b) $40$.

(c) $-3/4$.

(d) $1$.

(e) $0.4055$ (or more precisely $\ln(3/2)$. You could evaluate this limit in different ways: e.g. using a Taylor expansion for small $\lambda$ or using l'Hôpital's rule. Let's provide a small sketch of both solutions:

1. For small $\lambda$, $a^{\lambda} \approx 1 + \lambda \ln a$. Thus $3^{\lambda} - 2^{\lambda} \approx \lambda (\ln 3 - \ln 2)$. Plug this into the expression and we readily obtain that the limit is $\ln(3/2)$.

2. We can define $f(\lambda) = 3^{\lambda} - 2^{\lambda}$ and $g(\lambda) = \lambda$. Then $f(0) = g(0) = 0$. L'Hôpital's rule now states that $\lim_{\lambda \to 0} \frac{f(\lambda)}{g(\lambda)} = \frac{f'(0)}{g'(0)}$. Evaluating this expression readily yields the desired limit.

L'Hôpital's rule will be one of the topics that you'll get familiar with during the bachelor programme. If you already want to read more about it, you can check out Section 7.12 in Essential Mathematics for Economic Analysis.

(f) $x^2 + 7x -8 = (x-1)(x+8)$, so $\lim_{x \to 1} \frac{x^2 +7x -8}{(x-1)} = \lim_{x \to 1} x+8 = 9$.

Question 7

Compute the following limits.

(a) $\lim_{x \to 0} \frac{x^3 + 3x^2 - 2x}{x}$

(b) $\lim_{h \to 0} \frac{(x+h)^3 - x^{3}}{h}$

(c) $\lim_{x \to 0} \frac{(x+h)^{3} - x^{3}}{h}$

(d) $\lim_{x \to 0} \frac{x^2 - 1}{x^2}$

(e) $\lim_{h \to 0} \frac{\sqrt{h+3} - \sqrt{3}}{h}$

(f) $\lim_{t \to -2} \frac{t^2 - 4}{t^2 + 10t + 16}$

(g) $\lim_{x \to 4} \frac{2 - \sqrt{x}}{4-x}$

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(a) $\lim_{x \to 0} x^2 + 3x -2 = -2$.

(b) $\lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} = \lim_{h \to 0} 3x^2 + 3hx + h^2 = 3x^2$.

(c) $\lim_{x \to 0} \frac{(x+h)^3 - x^3}{h} = \lim_{h \to 0} 3x^2 + 3hx + h^2 = h^2.$

(d) $\lim_{x \to 0} 1 - \frac{1}{x^2} = -\infty$, in other words, the limit does not exist.

(e) $\lim_{h \to 0} \frac{\sqrt{h + 3} - \sqrt{3}}{h} \times \frac{\sqrt{h + 3} + \sqrt{3}}{\sqrt{h + 3} + \sqrt{3}} = \lim_{h \to 0} \frac{1}{\sqrt{h+3} + \sqrt{3}} = \frac{1}{2\sqrt{3}}$. Or equivalently, $\sqrt{3}/6$.

(f) Note that $t^2 - 4 = (t+2)(t-2)$ and $t^2 + 10t +16 = (t+8)(t+2)$. So an equivalent expression is $\lim_{t \to -2} \frac{t-2}{t+8} = -4/6 = -2/3$.

(g) Note that $4 - x = (2 - \sqrt{x})(2 + \sqrt{x})$. So we can write the exercise equivalently as $\lim_{x \to 4} \frac{1}{2\sqrt{x}} = 1/4$.

Question 8

Compute the derivatives of the following functions.

(a) $y = 9x^{10}$

(b) $y = \pi^{7}$

(c) $y = -4x^{-7}$

(d) $y = \frac{-2}{x^{2}}$

(e) $y = \frac{3}{\sqrt[3]{x}}$

(f) $y = \frac{-2}{x\sqrt{x}}$

Show answer

(a) $90x^9$.

(b) $0$.

(c) $28x^{-8}$.

(d) $4x^{-3}$.

(e) $-x^{-4/3}$.

(f) $3x^{-5/2}$.

Question 9

Suppose we know $g'(x)$. Find expressions for the derivatives of the following:

(a) $2g(x) + 3 \qquad$(b) $\frac{g(x) - 5}{3}$

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(a) $2g' (x)$.

(b) $\frac{1}{3} g'(x)$.

Question 10

For each of the following functions, find a function $F(x)$ having $f(x)$ as its derivative - that is, find a function that satisfies $F'(x) = f(x)$.

(a) $f(x) = x^{2} \qquad$(b) $f(x) = 2x + 3 \qquad$(c) $f(x) = x^{a}$, for $a \neq -1$

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(a) $F(x) = \frac{1}{3}x^3 + C$.

(b) $F(x) = x^2 + 3x + C.$

(c) $F(x) = \frac{1}{a+1}x^{a+1} + C.$

Question 11

The following limits are all of the form $\lim_{h \to 0} [f(a+h) - f(a)]/h$. Use your knowledge of derivatives to find the limits.

(a) $\lim_{h \to 0} \frac{(5+h)^{2} - 5^{2}}{h} \qquad$(b) $\lim_{s \to 0} \frac{(s+1)^{5} - 1}{s} \qquad$(c) $\lim_{h \to 0} \frac{5(x+h)^{2} + 10 - 5x^{2} - 10}{h}$

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(a) With $f (x) = x^2$ and $a = 5$, one has $\lim_{h \to 0} \frac{(5 + h)^{2} − 5^2}{h} = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a) = f'(5)$. Since $f'(x) = 2x$, we have $f'(5) = 10$ and thus the limit is $10$.

(b) Let $f (x) = x^5$. Then $f' (x) = 5x^4$ , and the limit is equal to $f'(1) = 5 \cdot 1^4 = 5$, based on the same argument as in (a).

(c) Let $f (x) = 5x^2 + 10$. Then $f'(x) = 10x$, and this is the value of the limit. (Note that $h$ goes to zero, not $x$. There is no specific value $a$).

Question 12

Differentiate the following functions with respect to $x$.

(a) $f(x) = 3x^{5} + 2x^{4} + 5$

(b) $f(x) = 8x^{4} + 2\sqrt{x}$

(c) $f(x) = (2x^2 - 1)(x^4 - 1)$

(d) $f(x) = \left( x^5 + \frac{1}{x} \right) (x^5 + 1)$

(e) $f(x) = x^{-1}(x^2 + 1)\sqrt{x}$

(f) $f(x) = \frac{1}{\sqrt{x^3}}$

(g) $f(x) = \frac{x+1}{x-1}$

(h) $f(x) = \frac{3x - 1}{x^2 + x + 1}$

(i) $f(x) = \frac{\sqrt{x} - 2}{\sqrt{x} + 1}$

(j) $f(x) = \frac{x^2 - 1}{x^{2} + 1}$

(k) $f(x) = \frac{ax + b}{cx + d}$

(l) $f(x) = x^{n} (a\sqrt{x} + b)$

Show answer

(a) $f'(x) = 15x^{4} + 8x^{3}$.

(b) $f'(x) = 32x^{3} + x^{-1/2}$.

(c) $f'(x) = 12x^{5} - 4x^{3} - 4x = 4x(3x^{4} - x^{2} -1)$.

(d) $f'(x) = 10x^9 + 5x^{4} + 4x^{3} - x^{-2}$.

(e) $f'(x) = \frac{3}{2}x^{1/2} - \frac{1}{2} x^{-3/2}$.

(f) $f'(x) = -\frac{3}{2} x^{-5/2}$.

(g) $f'(x) = -\frac{2}{(x-1)^{2}}$.

(h) $f'(x) = \frac{-3x^2 + 2x + 4}{(x^2 + x + 1)^{2}}$.

(i) $f'(x) = \frac{3}{2\sqrt{x}(\sqrt{x} + 1)^2}$.

(j) $f'(x) = \frac{4x}{(x^2 + 1)^{2}}$.

(k) $f'(x) = \frac{ad - bc}{(cx + d)^{2}}$.

(l) $f'(x) = a\left( n + \frac{1}{2} \right) x^{n-1/2} + nbx^{n-1}$.

Question 13

For the following functions, determine the intervals on which it is increasing.

(a) $y = \frac{1}{4}(x^{4} - 6x^{2})$

(b) $y = \frac{2x}{x^{2} + 2}$

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(a) $\left[ -\sqrt{3}, 0 \right]$ and $\left[\sqrt{3}, \infty \right)$.

(b) $\left[ -\sqrt{2}, \sqrt{2} \right]$.

Question 14

Find the equations for the tangents to the graphs of the following functions at the specified points.

(a) $y = 3 - x - x^2$ at $x =1$

(b) $y = \frac{x^2 - 1}{x^2 + 1}$ at $x = 1$

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(a) $y = -3+4$.

(b) $y = x-1$.

Question 15

If $f(x) = \sqrt{x}$, then $f(x) \cdot f(x) = x$. Differentiate this equation using the product rule in order to find a formula for $f'(x)$.

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The product rule yields $f'(x) \cdot f (x) + f(x) \cdot f'(x) = 1$, so $2f'(x) \cdot f (x) = 1$. Hence, $f '(x) = 1/2f (x) = 1/2\sqrt{x}$.