What is the derivative of #sinxcosx#?

Question

Emma Aldridge · Accepted Answer

#d/dx(sinxcosx) = cos2x#

The product rule can be used to differentiate any function of the form #f(x) = g(x)h(x)#. It states that #color(red)(f'(x) = g'(x)h(x) + g(x)h'(x)#.

The derivative of #sinx# is #cosx# and the derivative of #cosx# is #-sinx#.

#f'(x) = cosx(cosx) + sinx(-sinx)#

#f'(x) = cos^2x - sin^2x#

Use the identity #cos2x = cos^2x - sin^2x#:

#f'(x) = cos2x#

Hopefully this helps!

Scarlett Allison · Answer

#cos2x.#

Let #y=sinxcosx=1/2(2sinxcosx)=1/2sin2x=1/2sinphi,#

where, #phi=2x.#

Here, we first need to know that, #d/dt{kf(t)}=kd/dt{f(t)},#

where, #k# is a costant.

#:. dy/dx=d/dx(1/2sinphi)=1/2d/dx(sinphi)#

We now use the Chain Rule, which states that :

If #y# is a function (fun.) of #phi, and, phi" is a fun. of "x# [ so

that #y# becomes a fun. of #x#], then, #dy/dx=dy/(dphi)(dphi)/dx#

#:. dy/dx=1/2d/dx(sinphi)#

#=1/2d/(dphi)(sinphi)((dphi)/dx)#

#=1/2cosphid/dx(2x)............[because, phi=2x]#

#=(1/2cosphi)(2d/dx(x))#

#=cosphi=cos2x#

Enjoy Maths!

Paisley Johnson · Answer

#cos2x#

Since this is the product of 2 functions, differentiate using the #color(blue)"product rule"#

#"Given " f(x)=g(x)h(x)" then"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=g(x)h'(x)+h(x)g'(x))color(white)(2/2)|)))to(A)#

The following #color(blue)"derivatives"# should be known.

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(sinx)=cosx" and " d/dx(cosx)=-sinx)color(white)(2/2)|)))#

here #g(x)=sinxrArrg'(x)=cosx#

#"and " h(x)=cosxrArrh'(x)=-sinx#

Enter these values in place of (A).

#f'(x)=sinx(-sinx)+cosx(cosx)#

#=cos^2x-sin^2x#

This may be simplified using the #color(blue)"trigonometric identity"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(cos2x=cos^2x-sin^2x)color(white)(2/2)|)))#

#rArrd/dx(sinxcosx)=cos2x#

Isabella Ackerman · Answer

undefined The derivative of sin(x) * cos(x) is cos(x)^2 - sin(x)^2.