Question

Question #836cd

Answer 1

`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)`