Leibniz Rule

Question

Apply the Leibniz differentiation rule to find the derivative of the accumulation function whose lower limit is zero and whose upper limit is x-squared with integrand sine of t.

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When the upper limit of an accumulation function is itself a function of x, the plain Fundamental Theorem is not enough and the Leibniz rule takes over by layering the chain rule on top. The governing formula is \frac{d}{dx}\int_0^{g(x)} f(t)\,dt = f(g(x))\cdot g'(x), where the integrand is evaluated at the moving boundary and then scaled by the boundary's rate of change. Here f(t) = \sin t and g(x) = x^2, so g'(x) = 2x. Substituting into the formula, the derivative equals \sin(x^2)\cdot 2x = 2x\sin(x^2). The chain-rule factor 2x is essential precisely because the upper limit varies nonlinearly with x. Option \sin(x^2) forgets the chain-rule multiplier coming from the variable upper limit. Option 2x\cos(x^2) wrongly differentiates the integrand into a cosine rather than merely evaluating it at the limit. Option \cos(x^2) commits both of those errors at once. This rule generalizes accumulation to moving boundaries, a recurring JEE Advanced configuration. As a final plausibility check, the 2x factor must appear because the upper limit grows quadratically, accelerating the accumulation of area, and the multiplier captures that acceleration consistently for all real x.

Leibniz Rule

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