Question

y= the integral from -1 to x of sin(t^2)dt find dy/dx

Answer 1

y'=0, because the integral from -1 to x of sin(t^2)dt should be a constant.

Answer 2

I need a break down, but that answer is wrong. The correct answer is -(1/2)x^(-1/2)sin(x)

Answer 3

Is the equation \[y=\int\limits_{x}^{-1}\sin(t^2)dt\]?

Answer 4

oops, the -1 on the bottom

Answer 5

switch the x and the -1

Answer 6

Wait up, let me solve this on paper.

Answer 7

Big oops. So, the answer comes from the Fundamental theorem of calculus part 1. It states that If f(x) is continuous on [a,b] then,
g(x)=int from a to x f(t)dt
g'(x)=f(x)
so then you just plug x into the sin(t^2)

Answer 8

and that's going to get me -(1/2)x^(-1/2)sin(x)?

Answer 9

How do you know that's the correct answer?

Answer 10

back of the book

Answer 11

In general, the FTC says
\[\frac{d}{dx}\left[\int_c^{g(x)}f(t)~dt\right]=f(g(x))\cdot g'(x)\]
where \(c\) is any constant.

The answer you suggest, @GriffinT, is not correct.

Answer 12

the book is a lie?

Answer 13

it doesn't say to use FTC

Answer 14

I beg to differ: http://www.wolframalpha.com/input/?i=D%5BIntegrate%5BSin%5Bt%5E2%5D%2C%7Bt%2C-1%2Cx%7D%5D

which is the exact same result you would get if you were to use the FTC.

Answer 15

aight book is wrong then i suppose

Answer 16

look at the alternate form though at bottom. looks similar to answer in book though :p
oh well ill just take one wrong, no biggy

Answer 17

The only alternate form I see is the exponential form of sine.