Differentiation questions (With detailed solutions)

Advanced Mathematics - Differentiation Solutions

Question 1

(a)(i) If \( x^m y^n = (x + y)^{m+n} \), show that \( \dfrac{dy}{dx} = \dfrac{y}{x} \)

Step 1: Take natural logarithm of both sides

\[ m\ln x + n\ln y = (m+n)\ln(x+y) \]

Step 2: Differentiate implicitly with respect to x

\[ \frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = \frac{m+n}{x+y}\left(1 + \frac{dy}{dx}\right) \]

Step 3: Collect like terms

\[ \left(\frac{n}{y} - \frac{m+n}{x+y}\right)\frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x} \]

Step 4: Simplify both sides

\[ \frac{nx - my}{y(x+y)}\frac{dy}{dx} = \frac{nx - my}{x(x+y)} \]

Step 5: Solve for dy/dx

\[ \frac{dy}{dx} = \frac{y}{x} \]

(a)(ii) Find \( \dfrac{dy}{dx} \) if \( (xy)^7 + 2(xy)^2 + 3 = \sqrt{x} \)

Step 1: Differentiate implicitly

\[ 7(xy)^6(y + x\frac{dy}{dx}) + 4(xy)(y + x\frac{dy}{dx}) = \frac{1}{2\sqrt{x}} \]

Step 2: Factor out common terms

\[ (y + x\frac{dy}{dx})[7(xy)^6 + 4xy] = \frac{1}{2\sqrt{x}} \]

Step 3: Solve for dy/dx

\[ \frac{dy}{dx} = \frac{\frac{1}{2\sqrt{x}[7(xy)^6 + 4xy]} - y}{x} \]

Final answer: \(\boxed{\dfrac{1 - 2y\sqrt{x}[7(xy)^6 + 4xy]}{2x^{3/2}[7(xy)^6 + 4xy]}}\)

(b) Show that \( \dfrac{dy}{dx} = \dfrac{2}{1 + x^2} \) if \( y = \cos^{-1}\left( \dfrac{1 - x^2}{1 + x^2} \right) \)

Step 1: Let \( x = \tan \theta \)

Then: \[ \frac{1 - x^2}{1 + x^2} = \cos 2\theta \] So \( y = \cos^{-1}(\cos 2\theta) = 2\theta \) (for \( \theta \in [0, \pi/2] \))

Step 2: Differentiate

\[ \frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx} = 2 \cdot \frac{1}{1 + x^2} \]

(c) Maclaurin expansion of \( e^x \sin x \)

Step 1: Find derivatives at x=0

\[ f(0) = 0 \] \[ f'(0) = e^0(\sin 0 + \cos 0) = 1 \] \[ f''(0) = e^0(2\cos 0) = 2 \] \[ f'''(0) = e^0(2\cos 0 - 2\sin 0) = 2 \]

Step 2: Construct series

\[ f(x) \approx 0 + x + \frac{2x^2}{2!} + \frac{2x^3}{3!} = x + x^2 + \frac{x^3}{3} \]

Step 3: Factor as required

\[ \frac{x}{3}(x^2 + 3x + 3) \]

Question 2

(a) Find \( \dfrac{dy}{dx} \) for \( y = \tan^{-1}\left[ \dfrac{x}{\sqrt{x^2 + 1}} \right] \)

Step 1: Let \( u = \dfrac{x}{\sqrt{x^2 + 1}} \)

Then \( y = \tan^{-1} u \)

Step 2: Chain rule

\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

Step 3: Compute du/dx using quotient rule

\[ \frac{du}{dx} = \frac{\sqrt{x^2+1} - x^2/\sqrt{x^2+1}}{x^2 + 1} = \frac{1}{(x^2 + 1)^{3/2}} \]

Step 4: Combine results

\[ \frac{dy}{dx} = \frac{1}{1 + \frac{x^2}{x^2+1}} \cdot \frac{1}{(x^2 + 1)^{3/2}} = \frac{1}{(x^2 + 1)^{3/2}} \]

Final answer: \(\boxed{\dfrac{1}{(x^2 + 1)^{3/2}}}\)

(b) Differentiate \( f(x) = \dfrac{1}{\sqrt{x}} \) from first principles

Step 1: Definition of derivative

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{1/\sqrt{x+h} - 1/\sqrt{x}}{h} \]

Step 2: Combine fractions

\[ = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}} \]

Step 3: Rationalize numerator

Multiply numerator and denominator by \( \sqrt{x} + \sqrt{x+h} \): \[ = \lim_{h \to 0} \frac{x - (x+h)}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} \]

Step 4: Simplify

\[ = \lim_{h \to 0} \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} = \frac{-1}{2x^{3/2}} \]

Final answer: \(\boxed{-\dfrac{1}{2x^{3/2}}}\)

Question 3

(a)(i) Find derivative of \( \sin\left( \dfrac{2x}{1 + x^2} \right) \) with respect to \( \tan^{-1}\left( \dfrac{2x}{1 - x^2} \right) \)

Step 1: Let \( u = \dfrac{2x}{1 + x^2} \) and \( v = \tan^{-1}\left( \dfrac{2x}{1 - x^2} \right) \)

Step 2: Recognize \( v = 2\tan^{-1}x \) (for \( |x| < 1 \))

Step 3: Compute derivatives

\[ \frac{du}{dx} = \frac{2(1 + x^2) - 2x(2x)}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2} \] \[ \frac{dv}{dx} = \frac{2}{1 + x^2} \]

Step 4: Apply chain rule

\[ \frac{d}{dv} \sin u = \frac{d\sin u/dx}{dv/dx} = \frac{\cos u \cdot du/dx}{dv/dx} = \frac{\cos u \cdot \frac{2(1-x^2)}{(1+x^2)^2}}{\frac{2}{1+x^2}} = \frac{(1-x^2)\cos u}{1+x^2} \]

Final answer: \(\boxed{\dfrac{(1 - x^2)}{1 + x^2} \cos\left( \dfrac{2x}{1 + x^2} \right)}\)

(a)(ii) If \( x = \cos^{-1} \alpha \), \( y = \ln x \), prove \( \dfrac{dy}{dx} = \tan x \)

Step 1: Express in terms of x

Given \( \alpha = \cos x \)

Step 2: Differentiate

\[ \frac{dy}{dx} = \frac{dy/d\alpha}{dx/d\alpha} = \frac{1/x}{-1/\sqrt{1 - \alpha^2}} = -\frac{\sqrt{1 - \cos^2 x}}{x} = -\frac{\sin x}{x} \]

Note: There appears to be an error in the original problem statement as the result doesn't match the given condition.

(b)(i) If \( m = e^{x+y} \cos(x-y) \), show \( \dfrac{\partial m}{\partial x} + \dfrac{\partial m}{\partial y} = 2m \)

Step 1: Compute partial derivatives

\[ \frac{\partial m}{\partial x} = e^{x+y}\cos(x-y) - e^{x+y}\sin(x-y) \] \[ \frac{\partial m}{\partial y} = e^{x+y}\cos(x-y) + e^{x+y}\sin(x-y) \]

Step 2: Add them together

\[ \frac{\partial m}{\partial x} + \frac{\partial m}{\partial y} = 2e^{x+y}\cos(x-y) = 2m \]

(b)(ii) Maclaurin expansion of \( \left(1 - \dfrac{x}{3}\right)^{1/2} \)

Step 1: General binomial expansion

\[ (1 + z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \cdots \]

Step 2: Apply to given function

\[ \left(1 - \frac{x}{3}\right)^{1/2} = 1 - \frac{x}{6} - \frac{x^2}{72} + \cdots \]

Step 3: Estimate \( \sqrt{6} \) using \( x = 1 \)

\[ \sqrt{\frac{2}{3}} \approx 1 - \frac{1}{6} - \frac{1}{72} \approx 0.8194 \quad (\text{Actual } \sqrt{6} \approx 2.449) \]

(c) Related rates problem

Given: Object rising at 120 m/s, observer 0.5 km away

Step 1: Setup

Let \( h \) = height, \( \theta \) = angle of elevation

\[ \tan \theta = \frac{h}{500}, \quad \frac{dh}{dt} = 120 \]

Step 2: Differentiate

\[ \sec^2 \theta \frac{d\theta}{dt} = \frac{1}{500} \frac{dh}{dt} \]

Step 3: At h = 500m

\[ \tan \theta = 1 \Rightarrow \sec^2 \theta = 2 \] \[ 2 \frac{d\theta}{dt} = \frac{120}{500} \Rightarrow \frac{d\theta}{dt} = 0.12 \text{ rad/s} \]

Final answer: \(\boxed{0.12 \text{ rad/s}}\)

Question 4

(a) If \( y = \cos^{-1}\left( \dfrac{1-x^2}{1+x^2} \right) \), show \( \dfrac{dy}{dx} = \dfrac{2}{1+x^2} \)

Step 1: Let \( x = \tan \theta \)

Then: \[ \frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta \] So \( y = \cos^{-1}(\cos 2\theta) = 2\theta \) (for \( \theta \in [0, \pi/2] \))

Step 2: Differentiate using chain rule

\[ \frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx} = 2 \cdot \frac{1}{1+x^2} \] since \( \frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2} \)

Result: \(\boxed{\dfrac{2}{1+x^2}}\)

(b) Inverted cone water tank problem

Given:
- Altitude = 20m, Base radius = 6m
- Water inflow rate = 3m³/min
- Find water level rise rate when depth = 8m

Step 1: Establish relationship between radius (r) and height (h)

By similar triangles: \[ \frac{r}{h} = \frac{6}{20} \Rightarrow r = \frac{3h}{10} \]

Step 2: Volume formula

\[ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{3h}{10}\right)^2 h = \frac{3\pi}{100}h^3 \]

Step 3: Differentiate with respect to time

\[ \frac{dV}{dt} = \frac{9\pi}{100}h^2 \frac{dh}{dt} \]

Step 4: Plug in known values

\[ 3 = \frac{9\pi}{100}(8)^2 \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{300}{576\pi} = \frac{25}{48\pi} \text{ m/min} \]

Final answer: \(\boxed{\dfrac{25}{48\pi} \text{ m/min}}\)

Question 5

(a) Trough water level problem

Given:
- Trough with isosceles triangular ends (width 5m, height 2m)
- Water pumped in at 6m³/s
- Find water height change rate when height = 1.2m

Step 1: Find width-height relationship

At any height h, width w satisfies: \[ \frac{w}{h} = \frac{5}{2} \Rightarrow w = \frac{5h}{2} \]

Step 2: Volume formula

\[ V = \frac{1}{2} \times \text{width} \times \text{height} \times \text{length} = \frac{1}{2} \times \frac{5h}{2} \times h \times 8 = 10h^2 \]

Step 3: Differentiate

\[ \frac{dV}{dt} = 20h \frac{dh}{dt} \]

Step 4: Solve for dh/dt

\[ 6 = 20(1.2) \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = 0.25 \text{ m/s} \]

Final answer: \(\boxed{0.25 \text{ m/s}}\)

(b)(i) Optimal circle radii problem

Given:
- Wire length \( 2\pi(0.12) = 0.24\pi \) m
- Form two circles with radii \( r_1 \) and \( r_2 \)
- Minimize sum of areas \( A = \pi r_1^2 + \pi r_2^2 \)

Step 1: Constraint equation

\[ 2\pi r_1 + 2\pi r_2 = 0.24\pi \Rightarrow r_1 + r_2 = 0.12 \]

Step 2: Express area in terms of one variable

\[ A = \pi r_1^2 + \pi (0.12 - r_1)^2 \]

Step 3: Find critical point

\[ \frac{dA}{dr_1} = 2\pi r_1 - 2\pi (0.12 - r_1) = 0 \Rightarrow r_1 = 0.06 \text{ m} \] Thus \( r_2 = 0.06 \) m

Final answer: \(\boxed{r_1 = r_2 = 0.06 \text{ m}}\)

(b)(ii) Taylor expansion of \( x^2 \ln x \)

Step 1: Compute derivatives at x=1

\[ f(1) = 0 \] \[ f'(x) = 2x\ln x + x \Rightarrow f'(1) = 1 \] \[ f''(x) = 2\ln x + 3 \Rightarrow f''(1) = 3 \] \[ f'''(x) = \frac{2}{x} \Rightarrow f'''(1) = 2 \] \[ f^{(4)}(x) = -\frac{2}{x^2} \Rightarrow f^{(4)}(1) = -2 \]

Step 2: Construct Taylor series

\[ f(x) \approx 0 + (x-1) + \frac{3}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 - \frac{2}{4!}(x-1)^4 \] \[ = (x-1) + \frac{3}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{12}(x-1)^4 \]

Final expansion: \(\boxed{(x-1) + \dfrac{3}{2}(x-1)^2 + \dfrac{1}{3}(x-1)^3 - \dfrac{1}{12}(x-1)^4 + \cdots}\)

Question 6

(a) Find derivative of \( xy + \sin(xy) = 1 \)

Step 1: Differentiate implicitly

\[ y + x\frac{dy}{dx} + \cos(xy)\left(y + x\frac{dy}{dx}\right) = 0 \]

Step 2: Factor common terms

\[ (1 + \cos(xy))\left(y + x\frac{dy}{dx}\right) = 0 \]

Step 3: Solve for dy/dx

Since \( 1 + \cos(xy) \neq 0 \) (because \( \sin(xy) = 1 - xy \neq 0 \)): \[ \frac{dy}{dx} = -\frac{y}{x} \]

Final answer: \(\boxed{-\dfrac{y}{x}}\)

(b) Ladder sliding problem

Given:
- 15m ladder
- Bottom slides at 2m/s
- Find angle change rate when \( \theta = \pi/3 \)

Step 1: Relate variables

Let x = distance from wall, θ = angle with wall: \[ \cos \theta = \frac{x}{15} \]

Step 2: Differentiate

\[ -\sin \theta \frac{d\theta}{dt} = \frac{1}{15} \frac{dx}{dt} \]

Step 3: Plug in values

At \( \theta = \pi/3 \), \( \sin(\pi/3) = \sqrt{3}/2 \), \( dx/dt = 2 \): \[ -\frac{\sqrt{3}}{2} \frac{d\theta}{dt} = \frac{2}{15} \Rightarrow \frac{d\theta}{dt} = -\frac{4}{15\sqrt{3}} \text{ rad/s} \]

Final answer: \(\boxed{-\dfrac{4}{15\sqrt{3}} \text{ rad/s}}\)

Question 7

(a) Police car chase problem

Given:
- Police car 0.6km north moving south at 60km/h
- SUV 0.8km east moving east at unknown speed
- Distance increasing at 20km/h

Step 1: Setup coordinates

Let police position be \( (0, y) \), SUV position be \( (x, 0) \) \[ y = 0.6 - 60t \quad \text{(since moving south at 60km/h)} \] \[ x = 0.8 + vt \quad \text{(v = unknown speed)} \]

Step 2: Distance formula

\[ D = \sqrt{x^2 + y^2} \] Differentiating: \[ \frac{dD}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{\sqrt{x^2 + y^2}} \]

Step 3: At t=0 \[ 20 = \frac{0.8v + 0.6(-60)}{1} \Rightarrow 20 = 0.8v - 36 \Rightarrow v = 70 \text{ km/h} \]

Final answer: \(\boxed{70 \text{ km/h}}\)

(b) Estimate \( \sqrt[4]{16.012} \) without calculator

Step 1: Linear approximation

Let \( f(x) = x^{1/4} \), near \( x = 16 \): \[ f(16.012) \approx f(16) + f'(16)(0.012) \]

Step 2: Compute derivative

\[ f'(x) = \frac{1}{4}x^{-3/4} \Rightarrow f'(16) = \frac{1}{4}(16)^{-3/4} = \frac{1}{32} \]

Step 3: Approximate

\[ f(16.012) \approx 2 + \frac{1}{32}(0.012) = 2 + 0.000375 = 2.000375 \]

Final estimate: \(\boxed{2.000375}\)