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Question 24 of 35

Q.If u=ex2u = e^{x^2}, then ∂u∂x\dfrac{\partial u}{\partial x} = ______ .

(a) 2ex22e^{x^2}
(b) 2x ex22x\,e^{x^2}
(c) 00
(d) ex2e^{x^2}
Puducherry TnboardTamil Nadu HSC First Year (DGE) Commerce Board 2023MCQ· 1mImportance★★★★★
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Applying the chain rule to u=ex2u = e^{x^2} gives ∂u∂x=2x ex2\dfrac{\partial u}{\partial x} = 2x\,e^{x^2}, so the answer is option (b).

Here u=ex2u = e^{x^2}. Differentiate partially with respect to xx using the chain rule ddxeg(x)=eg(x) g′(x)\dfrac{d}{dx}e^{g(x)} = e^{g(x)}\,g'(x) with g(x)=x2g(x) = x^2:

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