Partial derivative of integral with variable limit

[Gregg]

Homework Statement

G(\theta, k ) = \int^{\theta}_0 g(x,k) dx

\frac{\partial G}{\partial \theta} = ?

\frac{\partial G}{\partial k} = ?

The Attempt at a Solution

If I say that \int g(x,k) dx = H(x,k)

\int^{\theta}_0 g(x,k) dx = H(\theta,k) - H(0,k)


Then is \frac{\partial G}{\partial \theta} = \frac{\partial H(\theta,k)}{\partial \theta}=g(\theta,k)
?

 

[vela]

Yup.

 

[HallsofIvy]

That is "correct" but your derivation is more complicated than necessary! By the fundamental theorem of calculus, the derivative with respect to \theta is just g(\theta, k).

Lagrange's formula is an extension of the fundamental theorem of calculus:
\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x, t) dt= f(x, \beta(x))}\frac{d\beta}{dx}- f(x, \alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x}dt.

Since \theta only appears in the upper limit of the integral, the derivative with respect to \theta is just
f(\theta, k)\frac{d\theta}{d\theta}= f(\theta, k)

and the derivative with respect to k is just
\int_0^\theta \frac{\partial f}{\partial k} dx

 

[Gregg]

That's clearer now, thank you.