Question #836cd

1 Answer
Jim G.
Mar 24, 2017

#dy/dx=mcos(mx)cos(nx)-nsin(mx)sin(nx)#

Explanation:

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

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

#color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=g(x)h'(x)+h(x)g'(x))color(white)(2/2)|)))#

#color(orange)"Reminder : " d/dx(sin(f(x))=cos(f(x)).f'(x)#

The same applies to #d/dx(cos(f(x)))=-sin(f(x)).f'(x)#

#"here " g(x)=sin(mx)rArrg'(x)=mcos(mx)#

#"and " h(x)=cos(nx)rArrh'(x)=-nsin(nx)#

#rArrdy/dx=sin(mx)(-nsin(nx)+cos(nx)(mcos(mx)#

#color(white)(rArrdy/dx)=mcos(mx)cos(nx)-nsin(mx)sin(nx)#

Related questions
See all questions in Chain Rule
Impact of this question
1468 views around the world
You can reuse this answer
Creative Commons License