Calculus S.M Yusuf Chapter 2 Ex 2.1 Solved Notes & Quiz

Question 1: Find the derivative of $y = f(x) = x^2 + 3x - 5$ from first principles (ab-initio method).

Solution:
Let $y = x^2 + 3x - 5$.
Step 1: Give increments $\Delta x$ to $x$ and $\Delta y$ to $y$:
$$y + \Delta y = (x + \Delta x)^2 + 3(x + \Delta x) - 5$$ Step 2: Subtract $y$ from both sides to find $\Delta y$:
$$\Delta y = \left[(x + \Delta x)^2 + 3(x + \Delta x) - 5\right] - (x^2 + 3x - 5)$$ $$\Delta y = x^2 + 2x\Delta x + (\Delta x)^2 + 3x + 3\Delta x - 5 - x^2 - 3x + 5$$ $$\Delta y = 2x\Delta x + (\Delta x)^2 + 3\Delta x$$ $$\Delta y = \Delta x (2x + \Delta x + 3)$$ Step 3: Divide both sides by $\Delta x$:
$$\frac{\Delta y}{\Delta x} = 2x + \Delta x + 3$$ Step 4: Apply the limit as $\Delta x \to 0$:
$$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} (2x + \Delta x + 3) = 2x + 0 + 3 = 2x + 3$$
Answer: $$\frac{dy}{dx} = 2x + 3$$

Question 2: Find the derivative of $f(x) = \frac{1}{\sqrt{x+1}}$ with respect to $x$ by definition.

Solution:
Let $y = (x + 1)^{-1/2}$.
Step 1: Apply increments:
$$y + \Delta y = (x + \Delta x + 1)^{-1/2}$$ $$\Delta y = (x + 1 + \Delta x)^{-1/2} - (x + 1)^{-1/2}$$ Step 2: Factor out $(x+1)^{-1/2}$:
$$\Delta y = (x + 1)^{-1/2} \left[ \left(1 + \frac{\Delta x}{x+1}\right)^{-1/2} - 1 \right]$$ Step 3: Apply the Binomial Theorem $(1 + t)^n = 1 + nt + \frac{n(n-1)}{2!}t^2 + \dots$ where $t = \frac{\Delta x}{x+1}$:
$$\Delta y = (x + 1)^{-1/2} \left[ 1 + \left(-\frac{1}{2}\right)\frac{\Delta x}{x+1} + \frac{(-1/2)(-3/2)}{2!}\left(\frac{\Delta x}{x+1}\right)^2 + \dots - 1 \right]$$ $$\Delta y = (x + 1)^{-1/2} \left[ -\frac{\Delta x}{2(x+1)} + \frac{3}{8}\frac{(\Delta x)^2}{(x+1)^2} - \dots \right]$$ Step 4: Divide by $\Delta x$ and take the limit $\Delta x \to 0$:
$$\frac{\Delta y}{\Delta x} = (x + 1)^{-1/2} \left[ -\frac{1}{2(x+1)} + \frac{3}{8}\frac{\Delta x}{(x+1)^2} - \dots \right]$$ $$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = (x + 1)^{-1/2} \cdot \left[ -\frac{1}{2(x+1)} + 0 \right]$$ $$\frac{dy}{dx} = -\frac{1}{2(x+1)^{1/2 + 1}} = -\frac{1}{2(x+1)^{3/2}}$$
Answer: $$\frac{dy}{dx} = -\frac{1}{2(x+1)^{3/2}}$$

Question 3: Find the derivative of $y = \sin(2x)$ from first principles.

Solution:
Let $y = \sin(2x)$.
Step 1: Apply increments:
$$y + \Delta y = \sin(2(x + \Delta x)) = \sin(2x + 2\Delta x)$$ $$\Delta y = \sin(2x + 2\Delta x) - \sin(2x)$$ Step 2: Use the trigonometric identity $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$:
Here $C = 2x + 2\Delta x$ and $D = 2x$:
$$\frac{C+D}{2} = \frac{4x + 2\Delta x}{2} = 2x + \Delta x, \quad \frac{C-D}{2} = \frac{2\Delta x}{2} = \Delta x$$ $$\Delta y = 2 \cos(2x + \Delta x) \sin(\Delta x)$$ Step 3: Divide by $\Delta x$:
$$\frac{\Delta y}{\Delta x} = 2 \cos(2x + \Delta x) \cdot \frac{\sin(\Delta x)}{\Delta x}$$ Step 4: Apply limit $\Delta x \to 0$ using fundamental limit $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$:
$$\frac{dy}{dx} = \lim_{\Delta x \to 0} \left[2 \cos(2x + \Delta x)\right] \cdot \lim_{\Delta x \to 0} \left[\frac{\sin(\Delta x)}{\Delta x}\right]$$ $$\frac{dy}{dx} = 2 \cos(2x + 0) \cdot 1 = 2 \cos(2x)$$
Answer: $$\frac{dy}{dx} = 2\cos(2x)$$

Question 4: Find the derivative of $y = x^{3/2}$ from first principles.

Solution:
Let $y = x^{3/2}$.
Step 1: Apply increments:
$$y + \Delta y = (x + \Delta x)^{3/2}$$ $$\Delta y = (x + \Delta x)^{3/2} - x^{3/2}$$ Step 2: Factor out $x^{3/2}$:
$$\Delta y = x^{3/2} \left[ \left(1 + \frac{\Delta x}{x}\right)^{3/2} - 1 \right]$$ Step 3: Expand using Binomial Theorem:
$$\Delta y = x^{3/2} \left[ 1 + \frac{3}{2}\left(\frac{\Delta x}{x}\right) + \frac{(3/2)(1/2)}{2!}\left(\frac{\Delta x}{x}\right)^2 + \dots - 1 \right]$$ $$\Delta y = x^{3/2} \left[ \frac{3}{2}\frac{\Delta x}{x} + \frac{3}{8}\frac{(\Delta x)^2}{x^2} + \dots \right]$$ Step 4: Divide by $\Delta x$ and evaluate limit:
$$\frac{\Delta y}{\Delta x} = x^{3/2} \left[ \frac{3}{2x} + \frac{3}{8}\frac{\Delta x}{x^2} + \dots \right]$$ $$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = x^{3/2} \left( \frac{3}{2x} \right) = \frac{3}{2} x^{3/2 - 1} = \frac{3}{2} x^{1/2}$$
Answer: $$\frac{dy}{dx} = \frac{3}{2}\sqrt{x}$$

Question 5: Differentiate $y = \frac{x^2 + 1}{x^2 - 1}$ with respect to $x$ using differentiation rules.

Solution:
Let $y = \frac{u(x)}{v(x)}$ where $u(x) = x^2 + 1$ and $v(x) = x^2 - 1$.
Step 1: Compute individual derivatives:
$$\frac{du}{dx} = \frac{d}{dx}(x^2 + 1) = 2x, \quad \frac{dv}{dx} = \frac{d}{dx}(x^2 - 1) = 2x$$ Step 2: Apply the Quotient Rule $\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$:
$$\frac{dy}{dx} = \frac{(x^2 - 1)(2x) - (x^2 + 1)(2x)}{(x^2 - 1)^2}$$ Step 3: Expand and simplify the numerator:
$$\frac{dy}{dx} = \frac{2x^3 - 2x - (2x^3 + 2x)}{(x^2 - 1)^2}$$ $$\frac{dy}{dx} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2}$$
Answer: $$\frac{dy}{dx} = -\frac{4x}{(x^2 - 1)^2}$$

Question 6: Differentiate $y = (2x + 3)^4 (3x - 1)^3$ with respect to