2 x sin(xt) If F(x) =∫--------dt, find F'(x) where x≠0 0 t Recall The Leibniz's Rule For Differentiation Of Integrals : Suppose f:[a,b]x[c,d]->R is continuous and f_x exists and is continuous on [a,b]x[c,d]. Let u(x) and v(x) be continuously differentiable functions from [c,d] into [a,b],and put v(x) F(x)=∫ f(t,x)dt,then : u(x) v(x) F'(x)= f(v(x),x)v'(x) -f(u(x),x)u'(x) +∫f_x(t,x)dt u(x) -------- In this problem : sin(xt) 2 Obviously , f(t,x)=--------- and f_x(t,x)=cos(xt) are both continuous on R t 2 ;u(x)=0 and v(x)=x are continuously differentiable on R so we can just apply this theorem to find F'(x) . 2 x F'(x)= f(v(x),x)v'(x) -f(u(x),x)u'(x) +∫f_x(t,x)dt 0 2 2 3 3 sin(x*x ) x 2sin(x ) 1 x =----------*2x- 0 + ∫cos(xt)dt =---------- + -∫cos(xt)d(xt) 2 0 x x 0 x 3 2sin(x ) 1 3 =--------- + -sin(x ) x x 3 3sin(x ) =------- x --------- sin(xt) 2 check: f(t,x)=--------- is continuous on R . t pf:It suffices to show that f(t,x) is continuous at (0,0) . The key point is : when |x| and |t| are small enough , |sin(xt)| <|xt|. sin(xt) sin(xt) xt ||(t,x)-(0,0)||<d , |---------- -0| =|-------| <|-----|=|x|<d . t t t -- ◢█◣ ◢█◣ ███ █ █ ε-δ method █ █ █ █ ◤█◥ █ █◢◣█ ██◢◣◢◣█ ██◢◤◢█◣ █ █◢◣ █ █◢◣ ◢█◣◢█◣ █ ██ ██ █ ██ ██ █▄◤ █ █ █ █ █ █ ◥▄█◥█◣ █ █◥█◤█ █ █◥█◤█ ◥▄◤ █ █ █ █ █ █ ◣ █◥█◤ ◇了解自己比認識別人重要◇對自己負責才能愛別人◇沒有多的事情◇ ◥▆◤㊣紅鳥 --

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