partial der

**thehollow89** · March 21st 2009, 02:37 PM

I am unsure where to begin with finding this partial derivative.
f(x,y)=integral from y to x of cos(t^2)dt

**mr fantastic** · March 21st 2009, 03:03 PM

Originally Posted by thehollow89

I am unsure where to begin with finding this partial derivative.
f(x,y)=integral from y to x of cos(t^2)dt

$f(x, y) = \int_y^x \cos (t^2) \, dt$ .

Using the Fundamental Theorem of Calculus and noting that y is treated as a constant when you differentiate with respect to x:

$\frac{\partial f}{\partial x} = \cos(x^2)$ .

Now note that $f(x, y) = \int_y^x \cos (t^2) \, dt = - \int_x^y \cos (t^2) \, dt$ and use the Fundamental Theorem of Calculus again (note that x is treated as a constant when you differentiate with respect to y):

$\frac{\partial f}{\partial y} = - \cos(y^2)$ .

**Danny** · March 21st 2009, 03:07 PM

Originally Posted by thehollow89

I am unsure where to begin with finding this partial derivative.
f(x,y)=integral from y to x of cos(t^2)dt

Depends on what partial derivative you want

$f(x,y) = \int_y^x \cos t^2\, dt = -\int_x^y \cos t^2 dt$

so using the 2nd fundemental theorem of Calculus

$\frac{\partial f }{\partial x } = \frac{\partial }{\partial x } \int_y^x \cos t^2 dt = \cos x^2$

$\frac{\partial f }{\partial y } = - \frac{\partial }{\partial y } \int_x^y \cos t^2 dt = - \cos y^2$

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