Ask a New Question
Menu

If y=3/(sinx+cosx) , find dy/dx

A. 3sinx-3cosx
B. 3/(sinx+cosx)^2
C. -3/(sinx+cosx)^2
D. 3(cosx-sinx)/(sinx+cosx)^2
E. 3(sinx-cosx)/(1+2sinxcosx)

1 answer

ever heard of the chain rule?

y = 3/u
y' = -3/u^2 u'

Now just plug in u = sinx+cosx
Similar Questions
  1. Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
    1. answers icon 1 answer
  2. tanx+secx=2cosx(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
    1. answers icon 0 answers
  3. I need to prove that the following is true. Thanks(cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds
    1. answers icon 0 answers
  4. prove the identity(sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1-cosX^2)^3 = (1-2CosX^2 + cos^4) (1-cosX^2)
    1. answers icon 0 answers
more similar questions