If f(x) = int_{x^2}^{x^4} sin sqrt{t} , dt,th | vedclass

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(B) To find $f'(x)$ for the integral $f(x) = \int_{g(x)}^{h(x)} \phi(t) \, dt$,we use the Leibniz integral rule:$f'(x) = \phi(h(x)) \cdot h'(x) - \phi

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$4x^3 \sin(x^2) - 2x \sin(x)$

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(B) To find $f'(x)$ for the integral $f(x) = \int_{g(x)}^{h(x)} \phi(t) \, dt$,we use the Leibniz integral rule:$f'(x) = \phi(h(x)) \cdot h'(x) - \phi(g(x)) \cdot g'(x)$Here,$\phi(t) = \sin \sqrt{t}$,$h(x) = x^4$,and $g(x) = x^2$.Calculating the derivatives:$h'(x) = \frac{d}{dx}(x^4) = 4x^3$$g'(x) = \frac{d}{dx}(x^2) = 2x$Substituting these into the formula:$f'(x) = \sin(\sqrt{x^4}) \cdot (4x^3) - \sin(\sqrt{x^2}) \cdot (2x)$Since $\sqrt{x^4} = x^2$ and $\sqrt{x^2} = |x|$,assuming $x > 0$ for simplicity in standard calculus problems,we get:$f'(x) = 4x^3 \sin(x^2) - 2x \sin(x)$Thus,the correct option is $B$.

If $f(x) = \int_{x^2}^{x^4} \sin \sqrt{t} \, dt$,then $f'(x)$ equals

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