Question

partial derivatives of:

Answer 1

\[f(x,y) = \int\limits_{y}^{x}\cos(t ^{2})dt\]

Answer 2

use the fundamental theorem of calculus

Answer 3

what's that?

Answer 4

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part

Answer 5

i still don't understand how to do it

Answer 6

\[\frac{\partial}{\partial x}\int\limits_{y}^{x}\cos(t ^{2})dt=\cos(x^2)\]

\[\frac{\partial}{\partial y}\int\limits_{y}^{x}\cos(t ^{2})dt=-\frac{\partial}{\partial y}\int\limits_{x}^{y}\cos(t ^{2})dt=-\cos(y^2)\]

Answer 7

let \[F(t)=\int f(t)dt\]

then \[\int\limits_{y}^{x}f(t)dt=F(x)-F(y)\]

thus \[\frac{\partial}{\partial x}\int\limits_{y}^{x}f(t)dt=\frac{\partial}{\partial x}(F(x)-F(y))=F'(x)=f(x)\]

and

\[\frac{\partial}{\partial y}\int\limits_{y}^{x}f(t)dt=\frac{\partial}{\partial y}(F(x)-F(y))=-F'(y)=-f(y)\]

Answer 8

so wrt x it's just cos(x^2)?

Answer 9

yes

Answer 10

and wrt y it's -cos(x^2)?

Answer 11

\[-\cos(y^2)\]

Answer 12

ah got cha! thanks!