Question

Find the derivative of these integrals....

Answer 1

\[d/dx (\int\limits_{-5}^{x} t^3/ (t^2 +1) dt) \]

Answer 2

\[d/dx \int\limits_{-3x}^{x^3} \sqrt{1+\sec(t^8)} dt\]

Answer 3

\[\frac{d}{dx} \int\limits_{a(x)}^{b(x)} g(t) dt \\ \text{ lets look at just the integral part and evaluate it like it is } \]
we will assume g(t) is continuous on [a,b]
\[I(x)=\int\limits_{a(x)}^{b(x)}g(t) dt =G(t)|_{a(x)}^{b(x)} \text{ assuming } G'=g \\ I(x)=G(b(x))-G(a(x)) \\ \text{ now we are asked \to differentiate this function I called } I \\ \frac{d}{dx}(I(x))=b'(x) G'(b(x))-a'(x)G'(a(x)) \text{ by chain rule } \\ \text{ remember } G'=g \text{ which was just the beginning integrand } \\ \frac{dI}{dx}=b'(x)g(b(x))-a'(x)g(a(x)) \]

Answer 4

basically you can just use that last thing I gave you as a formula

Answer 5

lier

Answer 6

why? @crazychickengirl123