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

1 Answer
Mar 5, 2016

Answer:

#(x+1)/(sqrt(x^2+2x-1))#

Explanation:

What we have here is a function within a function; #x^2+2x-1# is under the radical (#sqrt()#) sign. That means we have to use the chain rule to differentiate, which says that you take the derivative of the "inside" function (in this case #x^2+2x-1#) and multiply it by the derivative of the whole function.

Begin by finding the derivative of #x^2+2x-1#. Using the power rule, the derivative is #2x+2#. Now onto the whole function. Note that we can write #sqrt(x^2+2x-1)# as #(x^2+2x-1)^(1/2)#. That means we can again apply the power rule:
#d/dx(x^2+2x-1)^(1/2) = 1/(2(x^2+2x-1)^(1/2))#

Now we can multiply this by the derivative of the inside function, which we found as #2x+2#. Performing this operation yields:
#1/(2(x^2+2x-1)^(1/2))*2x+2 = (2x+2)/(2(x^2+2x-1)^(1/2))#

Finally, look for any ways to simplify the problem. We see that there is a #2# in the denominator - is there any way we can get rid of it? In fact, there is by factoring out a #2# from the numerator; take a look:
#(2(x+1))/(2(x^2+2x-1)^(1/2)) = (x+1)/((x^2+2x-1)^(1/2))#

Because the problem was given to us in radical form, we should convert it back, rewriting the answer as:
#(x+1)/(sqrt(x^2+2x-1))#