let-F-x-u-x-v-x-f-x-t-dt-how-to-calculate-dF-dx-x-

Question Number 64355 by turbo msup by abdo last updated on 17/Jul/19

l e t F (x) = \int_{u (x)}^{v (x {} f (x, t) d t

h o w t o c a l c u l a t e \frac{d F}{d x} (x) ?

Commented by MJS last updated on 17/Jul/19

I believe since we {don}^{'} t know the “ {status}^{″} of

the constant factor x in \int f (x, t) d t we cannot

give a general formula other than this one :

G (x) = \underset{u (x)}{\int^{v (x)}} f (t) d t = F (v (x)) - F (u (x)) + C

\frac{d G}{d x} = F^{'} (v (x)) v^{'} (x) - F^{'} (u (x)) u^{'} (x)

trying these simple examples I could not see

a general formula

f (x, t) = x + t

G (x) = \frac{v^{2}}{2} + v x - \frac{u^{2}}{2} - u x

G^{'} (x) = (v + x) v^{'} - (u + x) u^{'}

f (x, t) = x t

G (x) = \frac{x}{2} (v^{2} - u^{2})

G^{'} (x) = x (v^{'} v - u^{'} u) + \frac{1}{2} (v^{2} - u^{2})

f (x, t) = \frac{x}{t}

G (x) = x \ln \frac{v}{u}

G^{'} (x) = \ln \frac{v}{u} + x (\frac{v^{'}}{v} - \frac{u^{'}}{u})

f (x, t) = t^{x}

G (x) = \frac{v^{x + 1} - u^{x + 1}}{x + 1}

G^{'} (x) = v^{x} (v^{'} + \frac{v \ln v}{x + 1} - \frac{v}{{(x + 1)}^{2}}) - u^{x} (u^{'} + \frac{u \ln u}{x + 1} - \frac{u}{{(x + 1)}^{2}})

f (x, t) = x^{t}

G (x) = \frac{x^{v} - x^{u}}{\ln x}

G^{'} (x) = x^{v - 1} (v^{'} x + \frac{v}{\ln x} - \frac{1}{\ln^{2} x}) - x^{u - 1} (u^{'} x + \frac{u}{\ln x} - \frac{1}{\ln^{2} x})

Commented by mathmax by abdo last updated on 17/Jul/19

t h a n k y o u s i r m j s f o r t h i s h a r d w o r k w h e n i f i n d a f o r m u l a

f o r t h i s i w i l l p o s t \dots