Question

How do you differentiate `y=2 csc x + 5 cos x`?

Answer 1

The derivative is `-5sinx-2cotxcscx`.

Explanation:

`color(white)=d/dx(2cscx+5cosx)`

`=d/dx(2cscx)+d/dx(5cosx)`

`=2*d/dx(cscx)+5*d/dx(cosx)`

`=2*d/dx(cscx)+5*-sinx`

`=2*d/dx(cscx)-5sinx`

(I'm going to switch some terms around just so it's easier to look at.)

`color(white)=2*d/dx(cscx)-5sinx`

`=-5sinx+2*d/dx(cscx)`

`=-5sinx+2*d/dx(1/sinx)`

Quotient rule:

`=-5sinx+2*((d/dx(1)*sinx-1*d/dx(sinx))/(sinx)^2)`

`=-5sinx+2*((0*sinx-1*d/dx(sinx))/sin^2x)`

`=-5sinx+2*((0*sinx-1*cosx)/sin^2x)`

`=-5sinx+2*((-cosx)/sin^2x)`

`=-5sinx+2*(-cosx/sinx*1/sinx)`

`=-5sinx+2*(-cotx*cscx)`

`=-5sinx-2cotxcscx`

That's the answer. Hope this helped!