DIFFERENTIATION

Solutions: 1. f is differentiable at a ⇒ f′(a) = lim_{x→a} \frac{f (x) − f (a)}{x − a} exists

⇒ lim_{x→a}(f (x) − f (a)) = lim_{x→a} \frac{f (x) − f (a)}{x − a}(x − a)

= lim_{x→a} \frac{f(x) − f(a)}{x − a} lim_{x→a}(x − a)

= f′(a) · 0

= 0

⇒ lim_{x→a} f (x) = f (a) ⇒ f is continuous at a.

2. (For example) f (x) =| x | Then f is continuous at 0 , but f is not differentiable at 0

3. f is differentiable at 1 ⇒ f is continuous at 1

⇒ lim_{x→1^{−}} ax^{2} + bx + 7 = lim_{x→1^{+}} 2bx + 2a

⇒ a + x + 7 = 2b + 2a ⇒ a + b = 7 ⋯ (1)

f is differentiable at 1

⇒ \{ \begin{matrix}f′(1) = \frac{d}{dx}(ax^{2} + bx + 7) |_{x=1} \\ = 2ax + b |_{x=1} \\ = 2a + b \\ f′(1) = \frac{d}{dx}(2bx + 2a) |_{x=1} \\ = 2b |_{x=1} \\ = 2b \end{matrix}

⇒ 2a + b = 2b ⇒ 2a − b = 0 ⋯ (2)

(1),(2) ⇒ \{ \begin{matrix}a + b = 7 \\ 2a − b = 0 \end{matrix}

⇒ a = \frac{7}{3} , b = \frac{14}{3}

4. (1)\frac{dy}{dx} = 3 sin^{2}(tan x) cos(tan x) sec^{2} x

(2)\frac{dy}{dx} = sec(cos^{\frac{1}{2}} (x^{2} + 5 − \frac{1}{x})) tan(cos^{\frac{1}{2}} (x^{2} + 5 − \frac{1}{x}))·

\frac{1}{2} cos^{− \frac{1}{2}} (x^{2} + 5 − \frac{1}{x}) · (− sin(x^{2} + 5 − \frac{1}{x})) · (2x + x^{−2})

(3)\frac{dy}{dx} = 2x cot(csc x) + x^{2}(− csc^{2}(csc x)(− csc x cot x))

+ sin x cos x + x cos x cos x + x sin x(− sin x) = 2x cot(csc x) + x^{2} csc^{2}(csc x) csc x cot x + sin x cos x + x cos^{2} x − x sin^{2} x

(4)\frac{dy}{dx} = (cos(x^{2} + 1)) · 2x + \frac{(x^{3} + 1) − (x + 2) · 3x^{2}}{(x^{3} + 1)^{2}}

= 2x cos(x^{2} + 1) + \frac{−2x^{3} − 6x^{2} + 1}{(x^{3} + 1)^{2}}

5. y′ = na_{n}x^{n−1} + (n − 1)a_{n−1}x^{n−2} + ⋯ + 3a_{3}x^{2} + 2a_{2}x + a_{1} y′′ = n(n−1)a_{n}x^{n−2}+(n−1)(n−2)a_{n−1}x^{x−3}+⋯+3·2a_{3}x+2a_{2} y^{(n)} = a_{n} · n! y^{(n+1)} = 0

6. y^{3} = x^{2} + cos \frac{x}{y}

⇒ \frac{d}{dx}(y^{3}) = \frac{d}{dx}(x^{2} + cos \frac{x}{y})

⇒ 3y^{2} \frac{dy}{dx} = 2x − (sin \frac{x}{y}) · \frac{y − x\frac{dy}{dx}}{y^{2}}

= 2x − (sin \frac{x}{y}) · \frac{1}{y} + (sin\frac{x}{y}) · \frac{x}{y^{2}} · \frac{dy}{dx}

⇒ (3y^{2} − \frac{x}{y^{2}} sin \frac{x}{y})\frac{dy}{dx} = 2x − \frac{1}{y} sin \frac{x}{y}

⇒ \frac{dy}{dx} = \frac{2x − \frac{1}{y} sin \frac{x}{y}}{2y^{2} − \frac{x}{y^{2}} sin \frac{x}{y}}

7. xy = sin x + cos y

⇒ \frac{d}{dx}(xy) = \frac{d}{dx}(sin x + cos y)

⇒ y + x\frac{dy}{dx} = cos x − sin y \frac{dy}{dx}

⇒ (x + sin y)\frac{dy}{dx} = cos x − y

·^{·}_{·} \frac{dy}{dx} = \frac{cos x − y}{x + sin y}

⇒ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dx}(\frac{cos x − y}{x + sin y})

= \frac{(− sin x − \frac{dy}{dx})(x + sin y) − (1 + cos y \frac{dy}{dx})(cos x − y)}{(x + sin y)^{2}}

=\frac{−x sin x − sin x sin y − cos x + y − (x + sin y + cos y cos x − y cos y)\frac{dy}{dx}}{(x + sin y)^{2}}

=\frac{−x sin x − sin x sin y − cos x + y − (x + sin y + cos y cos x − y cos y)\frac{cos x − y}{x + sin y}}{(x + sin y)^{2}}

8.

(1) \{ \begin{matrix}2^{4} − 4 · 2^{2} = 0 \\ 3^{4} − 9 · 3^{3} = 0 \end{matrix}

⇒ 2^{4} − 4 · 2^{2} = 3^{4} − 9 · 3^{2} ⇒ (3,2) is on the curve C. Similarly, for (3,-2).

(2) y^{4} − 4y^{2} = x^{4} − 9x^{2}

⇒ \frac{d}{dx}(y^{4} − 4y^{2}) = \frac{d}{dx}(x^{4} − 9x^{2})

⇒ 4y^{3} \frac{dy}{dx} − 8y \frac{dy}{dx} = 4x^{3} − 18x

⇒ \frac{dy}{dx} = \frac{4x^{3} − 18x}{4y^{3} − 8y}

=\{ \begin{matrix}\frac{27}{8}  (x, y) = (3, 2) \\ −\frac{27}{8}  (x, y) = (3, −2) \end{matrix}

·^{·}_{·} Equation : y − 2 = \frac{27}{8}(x − 3)

(3) slope of the tangent · slope of the normal = -1 ⇒ −\frac{27}{8} · slope of the normal = -1

⇒ slope of the normal = \frac{8}{27}

·^{·}_{·} Equation : y + 2 = \frac{8}{27}(x − 3)

9. (1) \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dy}{dt}}

= \frac{sin t}{1 − cos t}

\frac{dy}{dx} |_{t= \frac{π}{3}} = \frac{sin \frac{π}{3}}{1 − cos \frac{π}{3}}

=\frac{\frac{\sqrt{3}}{2}}{1 − \frac{1}{2}}

=\sqrt{3}

\frac{d^{2}y}{dx^{2}} = \frac{d(\frac{dy}{dx})}{dx}

=\frac{dy′}{dx}

= \frac{\frac{dy′}{dt}}{\frac{dx}{dt}}

= \frac{\frac{cos t(1 − cos t) − sin t · sin t}{(1 − cos t)^{2}}}{1 − cos t}

=\frac{cos t − 1}{(1 − cos t)^{2}} · \frac{1}{1 − cos t}

= - \frac{1}{(1 − cos t)^{2}}

\frac{d^{2}y}{dx^{2}} |_{t= \frac{π}{3}} = −\frac{1}{(1 − cos t)^{2}}

=−\frac{1}{4}

(2) t= \frac{π}{3}

x = \frac{π}{3} − sin \frac{π}{3} = \frac{π}{3} − \frac{\sqrt{3}}{2}

y = 1 − cos \frac{π}{3} = \frac{1}{2}

Equation : y − \frac{1}{2} = \sqrt{3}(x − \frac{π}{3} + \frac{\sqrt{3}}{2})

10. Let x = distance between A and the intersection y = distance between B and the intersection s = distance between A and B

Then \{ \begin{matrix}tan θ = \frac{x}{y} \\ s^{2} = x^{2} + y^{2} \end{matrix}

Now \frac{dx}{dt} = - 3 m / sec , \frac{dy}{dt} = 2 m / sec

tan θ = \frac{x}{y}

⇒ \frac{d}{dt} tan θ = \frac{d}{dt}(\frac{x}{y})

⇒ sec^{2} θ\frac{dθ}{dt} = \frac{\frac{dx}{dt} · y − x\frac{dy}{dt}}{y^{2}}

x = 10 m , y = 25 m

⇒ tan θ = \frac{10}{25} = \frac{2}{5}

⇒ sec^{2} θ = \frac{29}{25}

·^{·}_{·} \frac{29}{25} \frac{dθ}{dt} = \frac{−3 · 25 − 10 · 2}{25^{2}}

\frac{dθ}{dt}= - \frac{19}{145} ( rad / sec ) ⋯ (1)

6 DIFFERENTIATION

s^{2} = x^{2} + y^{2}

⇒ \frac{d}{dt}s^{2} = \frac{d}{dt}(x^{2} + y^{2})

⇒ 2s\frac{ds}{dt} = 2x\frac{dx}{dt} + 2y \frac{dy}{dt}

x = 10 m , y = 25 m ⇒ s = \sqrt{10^{2} + 25^{2}} m

= 5 \sqrt{29} m

·^{·}_{·} 2 · 5\sqrt{29} · \frac{ds}{dt} = 2 · 10 · (−3) + 2 · 25 · 2

\frac{ds}{dt} = \frac{4}{\sqrt{29}} ( m / sec ) ⋯ (2)

11. (1) f(x) = x^{\frac{1}{3}}

⇒ f′(x) = \frac{1}{3}x^{− \frac{2}{3}}

⇒ df = f′(x)dx

=\frac{1}{3}x^{− \frac{2}{3}} dx

(2) Let f(x) = \sqrt[3]{x} Note f(x + Δx) ≈ f(x) + f′(x)Δx (Δ x is small) Take x = 1000 , Δx = 0.2

f(1000.2) ≈ f(1000) + f′(1000) · 0.2

= 10 + \frac{1}{3} · \frac{1}{100} · 0.2

≈ 10.0006

·^{·}_{·} \sqrt[3]{1000.2} ≈ 10.0006