Methods of Differentiation encompasses all techniques for computing derivatives, a topic carrying 2-3 questions per year in JEE Main. The derivative f'(x) measures the instantaneous rate of change of f at x and geometrically represents the slope of the tangent line.

The foundation rests on six differentiation rules: the power rule (d/dx($x^n$) = nx^(n-1)), constant multiple rule, sum/difference rule, product rule (d/dx(fg) = f'g + fg'), quotient rule (d/dx$\frac{f}{g}$ = $\frac{f'g - fg'}{g}$^2), and the chain rule (d/dx(f(g(x))) = f'(g(x))*g'(x)).

The chain rule is the most critical and frequently tested rule. For nested compositions, apply it layer by layer from the outermost function inward: d/dx[f(g(h(x)))] = f'(g(h(x))) * g'(h(x)) * h'(x). The most common mistake is forgetting the inner derivative.

Standard derivatives must be memorized perfectly: all six trigonometric functions, their inverses, exponentials, and logarithms. A useful pattern is that all "co-function" derivatives (cos, cot, cosec, and their inverses) carry a negative sign.

For inverse trigonometric differentiation, the JEE strategy is to SIMPLIFY BEFORE DIFFERENTIATING. Recognizing patterns like 2$\frac{x}{1-x^2}$ = tan(2tan^(-1)(x)) or 2xsqrt(1-$x^2$) = sin(2sin^(-1)(x)) converts complex chain rule problems into trivial derivatives. However, the domain of x critically affects the sign: for sin^(-1)(2xsqrt(1-$x^2$)), the derivative is +2/sqrt(1-$x^2$) when |x| < 1/sqrt(2) but -2/sqrt(1-$x^2$) when x > 1/sqrt(2).

Implicit differentiation handles equations like $x^3$ + $y^3$ = 3axy where y cannot be easily isolated. Differentiate both sides with respect to x, apply chain rule to y-terms $\frac{each gets a dy}{dx factor}$, then solve for dy/dx algebraically.

Parametric differentiation gives dy/dx = $\frac{dy/dt}{(dx/dt)}$ for curves defined parametrically. The second derivative formula is $d^{2y}$/$dx^2$ = [d/dt$\frac{dy}{dx}$]/$\frac{dx}{dt}$. A critical trap: $d^{2y}$/$dx^2$ is NOT $\frac{d^2y/dt^2}{(d^2x/dt^2)}$.

Logarithmic differentiation is essential for f(x)^g(x) forms where both base and exponent are variable. Take ln of both sides, differentiate implicitly, then multiply by y. This also simplifies products of many functions.

Higher-order derivatives use Leibniz's theorem: (uv)^(n) = sum C(n,r)*u^(n-r)*v^(r), analogous to the binomial theorem with derivatives.