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

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Answer 1 · 2015-06-01T19:47:50

This expression can be rewritten as #2(x+1)^-1#, following the exponential alw that states #a^-n=1/a^n#.

Naming #u=x+1#, we can rewrite the expression as #y=2u^-1# and, thus, derivate it according to the chain rule, which states that

#(dy)/(dx)=(dy)/(du)(du)/(dx)#

So,

#(dy)/(du)=-2u^-2#

#(du)/(dx)=1#

Thus

#(dy)/(dx)=(-2u^-2)(1)=-2(x+1)^-2=color(green)(-2/(x+1)^2)#