How do you find the derivative of # y = (2cosx)/(x+1)#?

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Answer 1 · 2017-03-01T15:53:17

#=((-2)(xsin(x)+sin(x)+cos(x)))/(x^2+2x+1)#

Recall that we use the quotient rule to differentiate a function of the form #(P(x))/(Q(x))#.

The quotient rule states that:
#d/(dx)(P(x))/(Q(x))=(Q(x)*P'(x)-P(x)*Q'(x))/(Q(x))^2#

I usually remember it as #(BT'-TB')/B^2# where #T# is the top function and #B# is the bottom function.

Now, applying the quotient rule to the function, we get:

#d/(dx)(2cosx)/(x+1)##=((x+1)(-2sinx)-(2cosx)(1))/(x+1)^2#

#=(-2xsin(x)-2sin(x)-2cos(x))/(x^2+2x+1)#

or
#=((-2)(xsin(x)+sin(x)+cos(x)))/(x^2+2x+1)#