AP Calculus AB and BC

Derivative of sqrt(x+1): Chain Rule

The derivative of the square root of x plus 1 is 1 over 2 times the square root of x plus 1. Rewrite the root as the power one half, apply the power rule, and the chain rule contributes a factor of 1 since the inside differentiates to 1.

ddx[x+1]=12x+1\frac{d}{dx}\left[\sqrt{x+1}\right] = \frac{1}{2\sqrt{x+1}}

Power rule plus chain rule

ddx(x+1)1/2=12(x+1)−1/2⋅1=12x+1\frac{d}{dx}(x+1)^{1/2} = \frac{1}{2}(x+1)^{-1/2}\cdot 1 = \frac{1}{2\sqrt{x+1}}

The inner derivative is 11, so the chain rule is invisible here. Change the inside to 2x+12x+1 and it stops being invisible: the derivative picks up a factor of 22.

The vertical tangent at the endpoint

The function is defined for x≥−1x \ge -1, but the derivative is not defined AT x=−1x = -1: the denominator is zero there and the slope runs to infinity.

So the graph starts at (−1,0)(-1, 0) with a vertical tangent and then flattens, which is the standard shape of every square root.

Common mistakes

  • Writing x+1=x+1\sqrt{x+1} = \sqrt{x}+1. At x=3x = 3 that claims 2.732.73 where the true value is 22.
  • Answering 12x+1\frac{1}{2}\sqrt{x+1}, multiplying by the half instead of dividing.
  • Claiming the derivative exists at x=−1x = -1. The function does; its derivative does not.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

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

It is 12x+1\frac{1}{2\sqrt{x+1}}.

What happens at x = -1?

The function is defined and equals 00, but the derivative is undefined: the tangent line is vertical.

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

It is 12x+1\frac{1}{\sqrt{2x+1}}, because the chain rule contributes a factor of 22 that cancels the 22 in the denominator.