Use the fundamental theorem of calculus to find the following derivative:\dfrac{d}{dx} \displaystyle \int_0^{x^2} \dfrac{1}{\sqrt{1+t^2}}\, dt.Choose the correct response from below.

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The problem asks us to differentiate an integral with an upper limit that depends on x: d/dx ∫ from 0 to x^2 of 1/√(1+t^2) dt.

First, recall the Fundamental Theorem of Calculus Part 1 with a variable upper limit: if F(x) = ∫ from a to g(x) f(t) dt, then F'(x) = f(g(x)) · g'(x).

Here, f(t) = 1/√(1+t^2) and g(x) = x^2, so F'(x) = f(g(x)) · g'(x) = (1/√(1+(x^2)^2)) · (d/dx x^2).

Compute the pieces: (x^2)^2 = x^4, so f(g(x)) = 1/√(1+x^4). The derivative of x^2 is 2x.

Multiply the two factors: F'(x) = (1/√(1+x^4)) · 2x = 2x / √(1+x^4).

Thus, the derivative evaluates to the expression 2x/√(1+x^4).

Use the fundamental theorem of calculus to find the following derivative:\dfrac{d}{dx} \displaystyle \int_0^{x^2} \dfrac{1}{\sqrt{1+t^2}}\, dt.Choose the correct response from below.

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