Differentiability

Because the variable appears in both the base and the exponent, neither the power rule nor the exponential rule applies directly, so logarithmic differentiation is the correct standard method. Take \ln y = x \ln x, then differentiate both sides implicitly: \frac{1}{y} \frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1, using the product rule on the right. Multiplying through by y = x^x gives \frac{dy}{dx} = x^x(1 + \ln x). Option x^x \ln x drops the contribution of differentiating the base, capturing only the exponential part. Option x \cdot x^{x-1} wrongly treats the exponent as a constant and applies the plain power rule. Option x^x(\ln x - 1) flips a sign in the product-rule term. The governing JEE Advanced technique is logarithmic differentiation for variable-base, variable-exponent functions, where treating either the base or the exponent as constant would silently discard half of the rate of change. Conceptually, the derivative splits into two contributions: one from how the base changes while the exponent is held fixed, and one from how the exponent changes while the base is held fixed, and logarithmic differentiation packages both cleanly through the term x \cdot (1/x) plus \ln x. This same product structure reappears in differentiating tower functions and in growth-rate problems, making the method a reusable JEE tool rather than a one-off trick. Plausibility check: at x = 1 the formula gives 1 \cdot (1 + 0) = 1, and the tangent slope of x^x at x = 1 is indeed 1, since the curve passes through (1,1) rising gently, confirming the derivative expression.