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

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Answer 1 · 2018-03-20T01:20:59

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

#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)#

#=-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!