Partial derivative of integral with variable limit

  • Thread starter Gregg
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  • #1
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Homework Statement



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

[tex] \frac{\partial G}{\partial \theta} = ? [/tex]

[tex] \frac{\partial G}{\partial k} = ? [/tex]

The Attempt at a Solution



If I say that [tex] \int g(x,k) dx = H(x,k) [/tex]

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


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

Answers and Replies

  • #2
vela
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Yup.
 
  • #3
HallsofIvy
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That is "correct" but your derivation is more complicated than necessary! By the fundamental theorem of calculus, the derivative with respect to [itex]\theta[/itex] is just [itex]g(\theta, k)[/itex].

Lagrange's formula is an extension of the fundamental theorem of calculus:
[tex]\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[/tex].

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

and the derivative with respect to k is just
[tex]\int_0^\theta \frac{\partial f}{\partial k} dx[/tex]
 
  • #4
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That's clearer now, thank you.
 

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