Methods of Differentiation
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Direct answer
Most JEE derivatives are not new derivatives — they are familiar ones strung together. The chain rule, dy/dx = dy/du × du/dx, handles nesting; the product rule d(uv)/dx = u'v + uv' and the quotient rule (u/v)' = (u'v − uv')/v^2 handle combinations; implicit differentiation works through an equation and solves for dy/dx; parametric differentiation divides dy/dt by dx/dt; and logarithmic differentiation (taking ln of both sides) tames both x^(sin x)-type towers and long products of powers. The method is chosen by the shape of the function before any computation begins — that choice is the actual skill being tested.
What you must remember
- Chain rule: d/dx f(g(x)) = f'(g(x)) × g'(x) — differentiate outer-to-inner and multiply; the most used rule in the paper.
- Product and quotient: (uv)' = u'v + uv'; (u/v)' = (u'v − uv')/v^2 — the quotient's minus sign is the classic slip.
- Implicit differentiation: differentiate both sides treating y as a function of x (so d/dx y^2 = 2y dy/dx), then isolate dy/dx.
- Parametric derivative: dy/dx = (dy/dt)/(dx/dt); the second derivative is d/dt(dy/dx) ÷ (dx/dt) — never (d^2y/dt^2)/(d^2x/dt^2).
- Logarithmic differentiation: for y = f(x)^g(x), write ln y = g(x) ln f(x), then differentiate; also the tool for products like (x + 1)^2 (x + 2)^3 (x + 3)^5.
- Inverse trigonometric chains: d/dx sin^(-1) x = 1/√(1 − x^2); d/dx tan^(-1) x = 1/(1 + x^2); composite forms usually need a trigonometric substitution first.
- Non-differentiable spots: |x| fails at 0 and the greatest-integer function fails at every integer — questions probe exactly these points.
Differentiating x to the power sin x
For y = x^(sin x), the power rule y' = sin x · x^(sin x − 1) is illegal, because both the base and the exponent vary. Take logarithms (the function demands x > 0 for a real value): ln y = sin x × ln x. Differentiate both sides, chain rule on the left: (1/y) y' = cos x × ln x + sin x × (1/x). Multiply through: y' = x^(sin x)[cos x ln x + sin x / x]. The structure of the answer is itself memorable — the original function times the derivative of the exponent-logarithm product — and it repeats for every variable-base-variable-exponent tower, from (sin x)^x to (ln x)^(cos x). A follow-up habit: state the domain before differentiating. For this function, x > 0; for (sin x)^x, the base must stay positive, which restricts x to intervals where sin x > 0. Examiners set the domain as a separate option precisely because the differentiation itself has been automated.
Where marks leak
JEE Main runs the machinery at speed: a three-layer chain, a parametric pair, a logarithmic derivative — one mark each, no tricks beyond the minus sign in the quotient rule. JEE Advanced probes the edges: differentiability of piecewise functions at the joining point (equate the one-sided derivatives, and check continuity first — continuity failing settles the question before derivatives are tried), and parametric second derivatives, where the division shortcut fails. The recurring losses: applying the power rule to variable exponents; computing d^2y/dx^2 as (d^2y/dt^2)/(d^2x/dt^2); and differentiating |x| blindly across 0. For implicit forms, forgetting that every y carries a dy/dx factor is the standard beginning of a lost solution.
Frequently asked questions
What is dy/dx for parametric equations?
(dy/dt)/(dx/dt), provided dx/dt ≠ 0 at the point of interest.
How do you differentiate x^(sin x)?
Logarithmically: y' = x^(sin x)(cos x ln x + sin x / x), for x > 0.
What is the second derivative in parametric form?
d/dt(dy/dx) divided by dx/dt — the naive ratio of second derivatives is wrong.
Where does |x| fail to be differentiable?
At x = 0, where the left and right slopes (−1 and 1) disagree.
What is the derivative of tan^(-1) x?
1/(1 + x^2), one of the inverse-trigonometric results used inside longer chains.