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

1 Answer
Feb 5, 2016

Answer:

#f'(x)=1/(2\sqrt{(x+1)}#

Explanation:

Given #f(x)=\sqrt{x+1}#. I'm sure you're familiar with the fact that if #y=x^n# then #dy/dx=nx^(n-1)# where #n# may be any real number and that if #g(y)# is a function on#f(x)# such that #g(f(x))#, then derivative of that is #g'(f(x))*f'(x))#

So that means we can rewrite the function as #f(x)=(x+1)^{1/2}# (using the law of indicies here).
So, differentiating it, #dy/dx=f'(x)=1/2(x+1)^(1/2-1)*frac{d}{dx}(x+1)=1/2(x+1)^(-1/2)#

So that's how I got the answer.