Question

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

Answer 1

`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.