Prove the following function is differentiable

**GIPC** · November 23rd 2010, 08:19 AM

Let f(t) be a continuous function under 1 variable, and we shall define

g(u,v)=Integral of f(t)dt from (v^2 - u^2) to (v^2+u^2)

prove that g(u,v) is differentiable in (u,v).

How do I approach? Should I use the Leibniz rule for integrals?

Thanks.

**roninpro** · November 23rd 2010, 06:23 PM

Is your integral this?

$\displaystyle g(u,v)=\int_{v^2-u^2}^{v^2+u^2} f(t)\text{ d}t$

Did you try using the Fundamental Theorem to compute $g_u$ and $g_v$ ?

Fundamental theorem of calculus - Wikipedia, the free encyclopedia

**HallsofIvy** · November 24th 2010, 04:29 AM

Which is, essentially, the Leibniz formula GIPC refers to.
$\frac{\partial g}{\partial u}= f(v^2+ u^2)(2u)- f(v^2- u^2)(-2u)$
$\frac{\partial g}{\partial v}= f(v^2+ u^2)(2v)- f(v^2- u^2)(2v)$
so not only do the partial derivatives exist but, since we are given that f is continuous, they are continuous functions and so g is differentiable.

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