How do you find the derivative of #f(x) = (sin x)(cos x)#?

2 Answers
Jul 21, 2016

Answer:

f'(x) = cos2x

Explanation:

differentiate using the #color(blue)"product rule"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(f(x)=g(x)h(x)rArrf'(x)=g(x)h'(x)+h(x)g'(x))color(white)(a/a)|)))#

#color(orange)"Reminder"#

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

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

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

Substitute these values into f'(x)

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

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

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

#rArrf'(x)=cos^2x-sin^2x=cos2x#
#color(magenta)"----------------------------------------------------------"#

Alternatively

#color(orange)"Reminder" color(red)(|bar(ul(color(white)(a/a)color(black)(sin2x=2sinxcosx)color(white)(a/a)|)))#

and recognise that #f(x)=sinxcosx=1/2sin2x#

differentiate f(x) using the #color(blue)"chain rule"#

#rArrf'(x)=1/2cos2x.d/dx(2x)=1/2cos2x.2=cos2x#

Jul 21, 2016

Answer:

#(d f(x))/(d x) (sin x)( cos x)=1-2 sin^2 x=cos 2x#

Explanation:

#(d f(x))/(d x) (sin x)( cos x)=?#

#d/(d x) sin x=cos x#

#d/(d x)cos x=-sin x#

#y=a*b" ; "y^'=a^'*b+b^'*a#

#(d f(x))/(d x) (sin x)( cos x)=cos x*cos x-sin x*sin x#

#(d f(x))/(d x) (sin x)( cos x)=cos^2 x-sin^2 x#

#"remember that :"cos^2 x=1-sin^2 x#

#(d f(x))/(d x) (sin x)( cos x)=1-sin^2 x-sin^2 x#

#(d f(x))/(d x) (sin x)( cos x)=1-2 sin^2 x=cos 2x#