Question

Find F ′(x) for F(x)=[2, x^3] sin(t^4) dt

Answer 1

\[\int\limits_{x^3}^{2} \sin (t^4) dt\]

Answer 2

You'd use the fundamental theorem of calculus. When you differentiate an integral, it effectively goes away. That's because integrals are the inverse of a derivative.

You'll be left with sin(t^4). Replace each 't' with x^3. Then simplify. Don't forget to multiply by the derivative of x^3 (basically using the chain rule)

Answer 3

Possible answers:
sin(2^4) – sin(x^12)
sin(x^7)
–sin(x^12)
–3x^2sin(x^12)

Answer 4

And the result will be negative since the x^3 portion is in the lower limit

Answer 5

Ok so
sin(t^4)
sin((x^3)^4) = sin(x^12)

Answer 6

that's part of it

Answer 7

then sin(x^12)*(3x)?

Answer 8

3x^2

Answer 9

yes, oops

Answer 10

–3x^2sin(x^12) would be the answer

if the x^3 was the upper limit, then the result wouldn't have a negative out front

Answer 11

Ok thanks so much!

Answer 12

sure thing

Answer 13

so first replace t with x^3 \[\sin((x^3)^4)\]
and then apply the chain rule ?
so why it is not \[\sin ( x^{12}) \cdot 12x^{11}\] ??

Answer 14

because we're applying the derivative to x^3

Answer 15

ohh okay so multiply the function by the derivative of x^3

Answer 16

ahh okay got it
not even applying the chain rule
thanks!