Question

find dy/dx given y = definite integral 1/(1+t^2). upper cosx. lower 0. show steps/explain.

Answer 1

\[\Large \frac{dy}{dx} \int_0^{\cos x} \frac{1}{1+t^2}dt \]
This one?

Answer 2

Seems like a problem for the fundamental theorem of Calculus.

Answer 3

well it says y = that integral. find dy/dx.

Answer 4

\[ \Large \int_0^x =F(x)-F(0)\]
If you derive the above you get again

\[ \Large \frac{dy}{dx} \int_0^x=f(x) \]

Answer 5

So I would say, just substitute cosx into t^2

Answer 6

ya, but what is the numerator

Answer 7

\[\Large \frac{dy}{dx} \int_0^{\cos x}=f(\cos x) = \frac{1}{1+\cos^2x}\]

Answer 8

that's the correct denom...

Answer 9

what's the solution then?

Answer 10

gotta have a -sinx in the top, eh?

Answer 11

im not sure... i think -sinx

Answer 12

hmm so I forgot about the dt.

Answer 13

but I wonder why that would not fall out of the solution anyway, because it's differentiated.

Answer 14

ah, I see. they want it like this?

\[\Large dy \int_0^{\cos x}=f(\cos x) dx \] ?

Answer 15

\[\Large \frac{d}{dx}\left(\int_0^{g(x)}f(t)dt\right)=\]\[\frac d{dx}\left(F(g(x))-\cancel{F(0)}^{\large0} \right)= F'(g(x))g'(x)=f(x)g'(x)\]chain rule

Answer 16

i dunno, but im satisfied with -sinx / 1+cos^2x

Answer 17

I repeat, chain rule
the derivative is with respect to x, and our upper bound is a function of x

Answer 18

oh that's where I went wrong.

Answer 19

ty both!

Answer 20

in general the fundamental theorem of calculus would be\[\Large \frac{d}{dx}\left(\int_{g(x)}^{h(x)}f(t)dt\right)\]\[\large =\frac d{dx}\left(F(h(x))-F(g(x))\right)= F'(h(x))h'(x)-F'(g(x))g'(x)\]where\[F'(x)=f(x)\]

Answer 21

\[ \Large \int_0^{\cos x} =F(\cos(x))-F(0) \]
Would have been best if I had made this 'extra' step to make sure about the chain rule

\[ \Large \frac{dy}{dx} \Large \int_0^{\cos x}=-f(\cos(x))\sin(x) \]

Answer 22

I wouldn't write the y in the dy/dx, since y is the integral itself and is already present in the equation

Answer 23

Yes you're right, standard Leibniz Notation?

\[ \Large \frac{d}{dx} \Large \int_{0}^{\cos x}f(t)dt=-f(\cos(x))\sin(x) \]

Answer 24

now that is exact, yeah

Answer 25

For the above case of course, the boundaries are depending on y.

Answer 26

okay thanks a lot!

Answer 27

welcome!