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derivadas de la funcion y= sen x

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3 Answers

Hola que tal,

La derivada que sale de  y =sen x, es la siguiente

y' = cos x 

Si necesitas alguna otra ayuda en ingles o en espanol te puedes poner en contacto conmigo yo estoy aqui para poder ayudarte en el idioma que usted escoja. 

atentatment, 

Yonathan c

y = sin x

dy/dx = y'(x)

then  y'(x)= cos (x) ======================================================================== Prove: ========================================================================= [1] Definition of sin (x) = [e^(xi)-e(-xi)] / 2 [2] y'(x)= y'(e^xi)/2 - y'(e^-xi)/2 ========================================================================= [3] y(x) = e(xi), y'(e^xi) = y'(x)= e^(xi) [4] y(x)= e^(-xi), y'(e^-xi) = y'(x)=-e^(-ix) then, substituing [3] and [4] on [2] have [5] y'(x) = e^(xi)/2 - -e^-xi)/2 ==> [6] y'(x) = e^(xi)/2 + e^(-xi)/2 ==> [7] y'(x) = [e^(xi) + e^(-xi)]/2 ========================================================================== Definition y(x)= cos (x) =[e^(xi) + e^(-xi)]/2 ========================================================================== then [8] y'(x) = [e^(xi) + e^(-xi)]/2 = cos (x) y'(x)= cos (x) That is, For y(x) = sin (x), y'(x)= cos (x)

Comments

[1].  A small exclusion is i in the denominator of definition of sin x.  Without it nothing works.

The y(x)=sin (x) definition is wrong, it should be:

y(x)=sin (x) = {e^(xi)-e^(-xi)}/2i;

let the operators d/dx = y'(x) = D[]

y'(x)= {D[e^(xi)]-D[e^(-xi)]}/2i, the

y'(x)= {ie^(xi)+ie^(-xi)}/2i = {e^(xi)+e^(-xi)}/2 = cos (x)

Then

for y(x)=sin (x) ==> y'(x)=cos (x)

 

Unfortunately, I only speak English and Russian.

However, it seems that you mean that you want the derivative of the sine function, y = sin x.

In that case the answer is y' = cos x.