Question

Take the derivative of
\[(d/dx )\int\limits_{a}^{\cos(x^(2)} \arcsin(\ln(t + 1))dt\]

would the answer be

arcsin(ln(-2xcos(x^(2))(sin(x^(2)) + 1)) ?
or
arcsin(ln(cos(x^(2)) + 1)*(-2xcos(x^(2))(sin(x^(2)) + 1))

Answer 1

b = cos(x^(2))
sorry about the mistake in latex

Answer 2

sorry is this the answer?
arcsin(ln(cos(x^(2)) + 1))*(-2xcos(x^(2))(sin(x^(2)) + 1))

Answer 3

Im assuming that it would be
b = cos(x^(2))

thus

arcsin(ln(b+1)*b')

Answer 4

\[\text{ Let } f(t)=\arcsin(\ln(t+1))\]
Let F be the antiderivative of f =>(F)'=f
So we have
\[\frac{d}{dx}[\int\limits_{a}^{\cos(\frac{x}{2})}f(t) dt ]=\frac{d}{dx}[ F(t)|_a^{\cos(\frac{x}{2})}]\]
\[\frac{d}{dx}[F(\cos(\frac{x}{2}))-F(a)]=\frac{d}{dx}F(\cos(\frac{x}{2}))-\frac{d}{dx}[F(a)]\]
\[\text{ apply chain rule } (\cos(\frac{x}{2}))'f(\cos(\frac{x}{2}))-0=(\cos(\frac{x}{2}))'f(\cos(\frac{x}{2})\]

Answer 5

\[\frac{1}{2}(-\sin(\frac{x}{2})) \cdot \arcsin(\ln(\cos(\frac{x}{2})+1))\]

Answer 6

Well looking at what @cuddlepony said I don't know if I have the right limits

Answer 7

you are suppose to ignore a

Answer 8

you are simply canceling the integral

Answer 9

lol oh you are the one asking the question
i'm so dumb

Answer 10

pff no you are not we all make mistakes

Answer 11

whats right? is arcsin(ln(b+1)*b')
the correct answer?

Answer 12

Well the problem I put down is a good example for yours

Answer 13

He is talking about the answer to the problem I thought we had

Answer 14

i mean the process.

Answer 15

the process is
you do a substitution of
b = cos(x^(2))

then take the derivative with respect to b
it cancels the integral

then you just apply chain rule to get

arcsin(ln(b+1)*b')

I'm confused as to where b' goes but i guess outside of the ln(b+1) seems correct

Answer 16

http://www.wolframalpha.com/input/?i=d%2Fdx%28integrate+x+dx+from+x+to+x%5E2%29