Example 21: $F(x) = \int_0^x \frac{\sin(xt)}{t} dt$ (for $x > 0$) Let $xt = u$, then $F(x) = \int_0^{x^2} \frac{\sin u}{(u/x)} \cdot \frac{du}{x} = \int_0^{x^2} \frac{\sin u}{u} du$ $F'(x) = \frac{\sin(x^2)}{x^2} \cdot 2x = \frac{2 \sin(x^2)}{x}$

9. Functional Equations and Differentiation

9.1 Standard Forms

Type 1: $f(x+y) = f(x) + f(y)$ (Cauchy's equation) If $f$ is differentiable at one point and satisfies this, then $f(x) = kx$. Proof: Differentiate with respect to $x$: $f'(x+y) = f'(x)$. This holds for all $x,y$, so $f'$ is constant. Thus $f(x) = kx + c$. From $f(0)= f(0+0) = f(0)+f(0) \Rightarrow f(0) = 0 \Rightarrow c = 0$.
Type 2: $f(x+y) = f(x)f(y)$ (Exponential type) If $f$ is differentiable and not identically zero, then $f(x) = a^x$. Proof: Differentiate w.r.t $x$: $f'(x+y) = f'(x)f(y)$. Put $x= 0$: $f'(y) = f'(0)f(y)$. Solve: $f(y) = C e^{f'(0)y}$. From $f(0)= 1 \Rightarrow C = 1$. Thus $f(x) = e^{kx}$ where $k = f'(0)$.
Type 3: $f(xy) = f(x) + f(y)$ (Logarithmic type) If $f$ is differentiable and not constant, then $f(x) = k \ln x$. Proof: Differentiate w.r.t $x$: $y f'(xy) = f'(x)$. Put $x= 1$: $y f'(y) = f'(1) \Rightarrow f'(y) = k/y$. Integrate: $f(y) = k \ln y + C$. From $f(1)= 0 \Rightarrow C = 0$.
Type 4: $f(x+y) = [f(x)+f(y)]/[1 - f(x)f(y)]$ (Tangent type) Solution: $f(x) = \tan(kx)$.

9.2 Method for Solving Functional Equations

1. Find $f(0)$ by suitable substitution
2. Find $f'(0)$ if possible
3. Differentiate the equation with respect to one variable
4. Solve the resulting differential equation
Example 22: $f(x+y) = f(x) + f(y) + 2xy - 1$, $f'(0) = 2$ Differentiate w.r.t $x$: $f'(x+y) = f'(x) + 2y$ Put $x = 0$: $f'(y) = f'(0) + 2y = 2 + 2y$ Integrate: $f(y) = 2y + y^2 + C$ From original: $f(0) = f(0+0) = f(0)+f(0)-1 \Rightarrow f(0) = 1$ Thus $C = 1$, so $f(x) = x^2 + 2x + 1 = (x+1)^2$.

10. Polar Coordinates

10.1 Derivatives in Polar Form

For curve $r = f(\theta)$: $x = r \cos \theta, y = r \sin \theta$ $\frac{dx}{d\theta} = \frac{dr}{d\theta} \cos \theta - r \sin \theta$ $\frac{dy}{d\theta} = \frac{dr}{d\theta} \sin \theta + r \cos \theta$ $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}$$ where $r' = dr/d\theta$.
Angle between tangent and radius vector: $$\tan \psi = \frac{r}{dr/d\theta}$$
Perpendicular from pole to tangent: $$p = \frac{r^2}{\sqrt{r^2 + (dr/d\theta)^2}}$$
Example 23: For cardioid $r = a(1 - \cos \theta)$, find $dy/dx$ $r' = a \sin \theta$ $$\frac{dy}{dx} = \frac{a \sin \theta \cdot \sin \theta + a(1-\cos \theta)\cos \theta}{a \sin \theta \cos \theta - a(1-\cos \theta)\sin \theta}$$ Simplify numerator: $a[\sin^2\theta + \cos \theta - \cos^2\theta] = a[\cos \theta - \cos 2\theta]$ Denominator: $a[\sin \theta \cos \theta - \sin \theta + \sin \theta \cos \theta] = a[2 \sin \theta \cos \theta - \sin \theta] = a \sin \theta (2 \cos \theta - 1)$ Thus $\frac{dy}{dx} = \frac{\cos \theta - \cos 2\theta}{\sin \theta (2 \cos \theta - 1)}$.

11. Proofs and Derivations

11.1 Derivative of $x^n$ from First Principles

$f(x) = x^n$ $f'(x) = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}$ Using binomial theorem: $(x+h)^n = x^n + \binom{n}{1} x^{n-1}h + \binom{n}{