Differentiation of multivariable function proof \frac{d}{dx} \int_{v(x)}^{u(x)} f(t,x)dt=u'(x)f(u(x),x)-v'(x)f(v(x),x)+\int_{v(x)}^{u(x)}\frac{\partial }{\partial x} f(t,x)dt

Walter Clyburn

Walter Clyburn

Answered question

2022-01-05

Differentiation of multivariable function proof
ddxv(x)u(x)f(t,x)dt=u(x)f(u(x),x)v(x)f(v(x),x)+v(x)u(x)xf(t,x)dt

Answer & Explanation

Melinda McCombs

Melinda McCombs

Beginner2022-01-06Added 38 answers

Start with
I(x)=vuf(t,x)dt=F(u,x)F(v,x)
Where F is the antiderivative of f
ddxI=ddxF(u,x)ddxF(v,x)
Now for a function w(a(x),b(x)) its derivative with respect to x can be written as
ax(x)wx(a,(x),b(x))+bx(x)wx(a,(x),b(x))
In terms of F we have F(u,x) and F(v,x) which remember u and v are functions of x. Therefore for each of them we can write this by plugging into the above formula. w(x)=F(x),a(x)=v(x) or u(x) and b(x)=x.
because xx=1 Now the above also holds for F(u,x) so
ddxF(u,x)=uxf(u,x)+xxf(u,x)
Therefore
I=f(v,x)+vxf(v,x)f(u,x)uxf(u,x)
Now rearranging you can see that it starts to take on your form.
I=vf(v,x)uf(u,x)+f(v,x)f(u,x)
Now the last two terms can be written in terms of an integral.
f(v,x)f(u,x)=uvf(t,x)dt
Which then can all come together to give
I=

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