西安毛子科技大学函数的求导法则XIDIANUNIVERSIT例8. 设 =f(x2),求 y解 y'= f(x2)-2x例9. y= f(cos? x)+ f(sin2 x), 求 y.解y'= f'(cos x) (cos? x) +f'(sin2 x) (sin2 x)= f(cos* x) -(2cosx) (-sin x)+f'(sin? x) (2sin x) (cosx)= -sin 2x· f'(cos2 x)+sin 2x· f'(sin2 x)= sin 2x·[f'(sin? x)- f'(cos? x)]函数的求导法则 例8. 设 求 2 y f x = ( ), y  . 解 y  = 2x 例9. 求 2 2 y f x f x = + (cos ) (sin ), y  . 解 2 2 y f x x    =  (cos ) (cos ) 2 = f x (cos ) 2 2 = −  +  sin 2 (cos ) sin 2 (sin ) x f x x f x   2 2 =  − sin 2 [ (sin ) (cos )] x f x f x   (2cos ) x  −( sin ) x 2 + f x (sin ) (2sin ) x (cos ) x 2 2 +  f x x   (sin ) (sin ) 2 f x ( )