What is the derivative of #1/(x+1) #?

1 Answer
Guilherme N.
Jun 1, 2015

This function can be rewritten as #(x+1)^-1#, following the law of exponentials that states #a^-n=1/a^n#.

Now, we can differentiate it using the chain rule, which, in turn, states that

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

In this case, if we rename #u=x+1#, then, we have that the original function #y=(x+1)^-1# becomes #y=u^-1# and, now, we can proceed to the chain rule steps:

#(dy)/(du)=-1*u^-2#

#(du)/(dx)=1#

Thus,

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

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