Answer to Question #174651 in Calculus for Leo Barrientos

Question

d/dx integral (top value=x^2, lower value=1) tan(t^3)dt

Answer 1

$\cfrac{d}{dx}(\int_{\alpha(x)}^{\beta(x)} f(t,x)dt) = \int_{\alpha(x)}^{\beta(x)}(\cfrac{\partial f}{\partial x}(t,x))dt + f(\beta(x),x) *\beta\prime(x) - f(\alpha(x), x)*\alpha\prime(x)$

$\cfrac{d}{dx}(\int_{1}^{x^2}\tan(t^3)dx) = \int_{1}^{x^2}\cfrac{\partial(\tan(t^3))}{\partial x}dx + 2x\tan(x^6) -\tan(1)*0=\\ =0 + 2x\tan(x^6) - 0 = 2x\tan(x^6)$