Section 3.5

Please post comments if you find any errors, or if you wish to contribute answers for additional problems. You may find information on using \LaTeX here.

1. (a) f'(x) = x^2\cos(x) + 2x\sin(x)

(c) g'(t) = -6t\sin(2t) + 3\cos(2t)

(e) f'(t) = -4\sin(3)t\sin(4t) + 3\cos(3t)\cos(4t)

(f) g'(z) = 12\sin^2(4z)\cos(4z)

2. (a) \displaystyle{\frac{dy}{dx} = \frac{2x\cos(2x) - \sin(2x)}{x^2}}

(c) \dfrac{dx}{dt} = 8t\cos(4t^2 + 1)

(e) \dfrac{dz}{dt} = 2\sec(2t)\tan(2t)

(g) \dfrac{dy}{dx} = -2x^2\csc(2x)\cot(2x) + 2x\csc(2x)

3. \dfrac{d}{dx}(\sin^2(2x)\cos^2(3x) = -6\sin^2(2x)\cos(3x)\sin(3x)
+ 4\cos^2(3x)\sin(2x)\cos(2x)

(c) \dfrac{d}{dq}\sec^3(q^2) = 6q\sec^3(q^2)\tan(q^2)

(e) \displaystyle{\frac{d}{dz}\sqrt{1 + \sin^2(z)} = \frac{\sin(z)\cos(z)}{\sqrt{1 + \sin^2(z)}}}

5. T(t) = 1

7. (a) S(x) = \dfrac{1}{2}x + 1

(b) T(x) = 4x

(c) U(x) = 2x + 1

(e) h = f \circ g and U = S \circ T

8. (a) \displaystyle{\lim_{x \to 0}\frac{\sin(2x)}{x} = 2}

(c) \displaystyle{\lim_{x \to 0}\frac{\tan(x)}{x} = 1}

(e) \displaystyle{\lim_{x \to 0}\frac{\sin^2(x)}{x} = 0}

(g) \displaystyle{\lim_{t \to 0}\frac{\sin^2(3t)}{t^2} = 9}

9. (a) f is O(h) , but not o(h) .

(c) g is O(h) , but not o(h) .

(e) f  is o(h) , and so also O(h) .

11. (a) D(0.00001) = 4.000010000027032 to 15 decimal places
D_1(0.00001) = 4.000000000026205 to 15 decimal places
D_2(0.00001) = 3.999999999848569 to 15 decimal places
f'(2) = 2

(c) D(0.00001) = 0.999999999983333 to 15 decimal places
D_1(0.00001) = 0.999999999983333 to 15 decimal places
D_2(0.00001) = 1.000000000000000 to 15 decimal places
f'(0) = 1

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