What is the derivative of #sqrt(x^2-1)#?

1 Answer
Feb 14, 2017

Answer:

#dy/dx=x/sqrt{x^2-1}#

Explanation:

Let's equate the function to a variable #y#, so that
#y=sqrt{x^2-1}#

Now, I'll take another variable #t# and equate it as such,
#t=x^2-1#

So that makes the #y# function as #y=sqrtt#

Now, we are to find the derivative of #y# with respect to #x#. So that means we are to find #dy/dx#

Now, we can use chain rule to simplify our problem as
#dy/dx=dy/dt*dt/dx#

That makes it, #dy/dx=d/dt(sqrtt)*d/dx(x^2-1)#

Now, we know that for any Real value #n#, #d/dx(x^n)=nx^(n-1)#
, and that the derivative of a constant is zero. So
#dy/dx=1/{2sqrtt}*2x#

Now, #t# was taken as #t=x^2-1#, so substituting that back into the equation gives us the answer being searched for (after a few simplifications of course).