find the derivative of ln x?

Question

If f(x)= ln (√x + 1) (on the the x in under the square root)

So I am not sure is it the answer

f'(x) = 1/(√x + 1)

Answer 1

We have f(x) = ln(sqrt(x) + 1). We know that, if u is a differentiable positive function of x, than the derivative of ln(u) is u'/u. Here, u(x) = sqrt(x) + 1, so that u'(x) = 1/(2sqrt(x)).

So, f'(x) = (1/(2sqrt(x)))/(sqrt(x) +1) = 1/(2*(x + sqrt(x)), that's the right answer

Answer 2

d(ln x)/dx = 1/x

In this case the ans is 1/(√x+1) * 1/(2√x)

You have to use chain rule. first find derivative of the term in brackets with log as if it was x. That gives your own answer. But then you have to multiply it with the derivative of the term within brackets which is what I've multiplied.

Answer 3

The answer is 1/(sqrt(x)+1) times 1/(2*sqrt(x))

Answer 4

You have the correct answer.