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

1 Answer
Feb 5, 2016

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 onf(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.