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NCERT Exemplar · Q72

Q.Derivative of x2x^2 w.r.t. x3x^3 is __________.

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The derivative of x2x^2 with respect to x3x^3 is found by treating x3x^3 as the independent variable. Using the chain rule in reverse, the result is 23x\frac{2}{3x}.

Concept and Intuition

When we say "derivative of yy with respect to uu", we mean dydu\frac{dy}{du} — the rate at which yy changes as uu changes. Here, y=x2y = x^2 and u=x3u = x^3. The catch is that both are functions of xx, not directly of each other. So we need a bridge.

The chain rule gives us exactly that bridge:

dydu=dy/dxdu/dx.\frac{dy}{du} = \frac{dy/dx}{du/dx}.

Think of it this way: if you know how yy changes with xx, and how uu changes with xx, then the ratio of those rates tells you how yy changes per unit change in uu. It’s like converting speeds: if a car travels 60 km per hour and its fuel gauge drops 5 litres per hour, then the fuel consumption is 560\frac{5}{60} litres per km.

Step-by-Step Solution

  1. Identify the functions

    We have y=x2y = x^2 and u=x3u = x^3. We want dydu\frac{dy}{du}.

  2. Differentiate each with respect to xx

dydx=2x,dudx=3x2.\frac{dy}{dx} = 2x, \quad \frac{du}{dx} = 3x^2.

  1. Apply the chain-rule formula

dydu=dy/dxdu/dx=2x3x2.\frac{dy}{du} = \frac{dy/dx}{du/dx} = \frac{2x}{3x^2}.

  1. Simplify Cancel one xx (provided x≠0x \neq 0): …

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