Question

Use part I of the Fundamental Theorem of Calculus to find the derivative of
y=∫sqrt(x) to -6 cos(t)/(t^7)dt

Answer 1

sqrt(x) is on the top and -6 on the bottom. sorry.

Answer 2

ok like this
$$ \Large { \int_{6}^{\sqrt{x}} \frac{\cos(t)}{t^7}~dt


}
$$

Answer 3

yeah

Answer 4

we can use the fundamental theorem of calculus
$$ \Large \rm {
\frac{d}{dx}\int_{a}^{x} f(t) dt = f(x)


}$$

Answer 5

but theres a slight hitch, the upper limit is sqrt(x), not x

Answer 6

we have to make a chain rule modification

Answer 7

oh ok. i put in cos(sqrt(x))/sqrtx^(7) but the computer marked me wrong

Answer 8

Fundamental theorem of calculus take two, (with chain rule)
$$ \Large \rm {
\frac{d}{dx}\int_{a}^{u(x)} f(t) dt = f(u(x)) \cdot u '(x)


}$$

Answer 9

did you mean f'(u(x))*u'(x) ?

Answer 10

$$
\Large {
\frac{d}{dx}\int_{6}^{\sqrt{x}} \frac{\cos(t)}{t^7}~dt
= \frac{\cos(\sqrt{x})}{(\sqrt{x})^7} \cdot \frac{1}{2\sqrt{x}}

}

$$

Answer 11

i meant f(u(x)) * u'(x)

Answer 12

here is an explanation with some examples
http://www.sosmath.com/calculus/integ/integ03/integ03.html

Answer 13

hmm for some reason they're marking it wrong. Did i plug it in wrong?

Answer 14

yes the square root x should be in the denominator

Answer 15

ok i see thank you!!

Answer 16

$$

\Large {
\frac{d}{dx}\int_{6}^{\sqrt{x}} \frac{\cos(t)}{t^7}~dt
= \frac{\cos(\sqrt{x})}{(\sqrt{x})^7} \cdot \color{blue}{\frac{1}{2\sqrt{x}}}

}

$$

Answer 17

yes that's the right answer