Question

дана функция y=sinx+cosx,вычислите y'(p/4)

Answer 1

у'=cosx -sinx,

y'(π/4) = cos(π/4) -sin(π/4) = (√2÷2) - (√2÷2) =0.

Ответ: 0.

Answer 2

$y=sin x+cos x=sqrt{2}(frac{sqrt{2}}{2}sinx+frac{sqrt{2}}{2}cos x)=sqrt{2}(sin x cos frac{pi}{4}+sin frac{pi}{4}cos x)=sqrt{2}sin (x+frac{pi}{4})$

 

$y'=(sqrt{2}sin(x+frac{pi}{4}))'=sqrt{2}*(sin(x+frac{pi}{4})'=sqrt{2}*cos(x+frac{pi}{4})*(x+frac{pi}{4})'=sqrt{2}*cos(x+frac{pi}{4})*1=sqrt{2}cos(x+frac{pi}{4})$

 

$y'(frac{pi}{4})=sqrt{2}cos(frac{pi}{4}+frac{pi}{4})=sqrt{2}*cos frac{pi}{2}=sqrt{2}*0=0$

 

либо так

$y'=(sin x+cos x)'=(sin x)'+(cos x)'=cos x+(-sin x)=cos x-sin x$

 

$y'(frac{pi}{4})=cos frac{pi}{4}-sin frac{pi}{4}=frac{sqrt{2}}{2}-frac{sqrt{2}}{2}$

ответ: 0