CalculusHigh Yield (2 Qs in JEE)

Differentiation & Methods Formulas

Product rule, quotient rule, chain rule, implicit differentiation, logarithmic differentiation, and higher derivatives.

7 Formulas•590+ Questions in Bank

Core Formulas & Mathematical Identities

\frac{d}{dx}(uv) = u'v + uv', \quad \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

Fundamental derivative rules for algebraic combinations.

Examiner Tip: Numerator of quotient rule starts with derivative of numerator: u'v.

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}, \quad \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \quad \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}

Derivative of composite functions and second derivative of parametric functions.

Examiner Tip: Do NOT just divide d²y/dt² by d²x/dt² for second derivative.

y = [u(x)]^{v(x)} \implies \frac{dy}{dx} = [u(x)]^{v(x)} \left[ v'(x)\ln(u(x)) + \frac{v(x)u'(x)}{u(x)} \right]

Differentiating variable-base to variable-exponent functions.

Examiner Tip: Standard result: d/dx (x^x) = x^x(1 + ln x).

\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}, \quad \frac{d}{dx}(\sec^{-1}x) = \frac{1}{|x|\sqrt{x^2-1}}

Derivatives of inverse trigonometric functions.

Examiner Tip: Derivative of co-functions (cos⁻¹x, cot⁻¹x, csc⁻¹x) has identical form with a minus sign.

\frac{d^n}{dx^n}(uv) = \sum_{r=0}^n {}^nC_r \frac{d^{n-r}u}{dx^{n-r}} \frac{d^r v}{dx^r}

Binomial-type expansion for calculating the n-th derivative of a product of two functions.

Examiner Tip: Set v as the polynomial term whose higher derivatives vanish after a few steps.

\frac{d}{dx} \begin{vmatrix} f & g \\ u & v \end{vmatrix} = \begin{vmatrix} f' & g' \\ u & v \end{vmatrix} + \begin{vmatrix} f & g \\ u' & v' \end{vmatrix}

Differentiate one row (or column) at a time, keeping all other rows unchanged, and sum the resulting determinants.

Examiner Tip: For 3×3 determinants, the result is the sum of 3 determinants corresponding to differentiating row 1, row 2, and row 3.

F(x, y) = 0 \implies \frac{dy}{dx} = -\frac{\partial F / \partial x}{\partial F / \partial y}

Fast evaluation using partial derivatives treating y as constant in numerator and x as constant in denominator.

Examiner Tip: Massive time saver in JEE for finding slopes of tangents to implicit algebraic curves.

Frequently tested misconceptions in Differentiation & Methods

•Applying d/dx(x^x) = x * x^(x-1): incorrect! Must use logarithmic differentiation to get x^x (1 + ln x).
•Parametric second derivative requires chain rule: d²y/dx² = (d/dt(dy/dx)) / (dx/dt).