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Derivative of x2+1\sqrt{x^{2} + 1}

The derivative of x2+1\sqrt{x^{2} + 1} is 2x2x2+1\frac{2 x}{2 \sqrt{x^{2} + 1}}.

f(x) =
ddx[x2+1]=2x2x2+1\frac{d}{dx}\left[\sqrt{x^{2} + 1}\right] = \frac{2 x}{2 \sqrt{x^{2} + 1}}

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Graph of f(x) = sqrt(x^2 + 1) and its derivative-3-2-1123-1123slope 0.7071f(x)f'(x)tangent at x = 1
Graph of f(x) = sqrt(x^2 + 1) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 0.70711.

How to work it out

  1. 1

    We differentiate f(x)=x2+1f(x) = \sqrt{x^{2} + 1} with respect to x. The rule that applies is the chain rule.

    f(x)=x2+1f(x) = \sqrt{x^{2} + 1}
  2. 2

    Chain rule: differentiate the outer function, keep the inside, then multiply by the derivative of the inside. The inside is g(x)=x2+1g(x) = x^{2} + 1.

    g(x)=x2+1,gā€²(x)=2xg(x) = x^{2} + 1,\quad g'(x) = 2 x
  3. 3

    The outer function is sqrt(u); its derivative evaluated at g(x), times g'(x):

    fā€²(x)=[12u]u=g(x)ā‹…gā€²(x)f'(x) = \left[\frac{1}{2 \sqrt{u}}\right]_{u=g(x)} \cdot g'(x)
  4. 4

    Simplify:

    fā€²(x)=2x2x2+1f'(x) = \frac{2 x}{2 \sqrt{x^{2} + 1}}
  5. 5

    Differentiating once more gives the second derivative, which tells you about curvature:

    fā€²ā€²(x)=4x2+1āˆ’2x22x2x2+1(2x2+1)2f''(x) = \frac{4 \sqrt{x^{2} + 1} - 2 x 2 \frac{2 x}{2 \sqrt{x^{2} + 1}}}{\left(2 \sqrt{x^{2} + 1}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\sqrt{x^{2} + 1}\right] = \frac{2 x}{2 \sqrt{x^{2} + 1}} \]
\[ f(x) = \sqrt{x^{2} + 1} \]
\[ g(x) = x^{2} + 1,\quad g'(x) = 2 x \]
\[ f'(x) = \left[\frac{1}{2 \sqrt{u}}\right]_{u=g(x)} \cdot g'(x) \]
\[ f'(x) = \frac{2 x}{2 \sqrt{x^{2} + 1}} \]
\[ f''(x) = \frac{4 \sqrt{x^{2} + 1} - 2 x 2 \frac{2 x}{2 \sqrt{x^{2} + 1}}}{\left(2 \sqrt{x^{2} + 1}\right)^{2}} \]

Values of f and f' for sqrt(x2x^{2} + 1)

xf(x)f'(x)
-22.236068-0.89442719
-11.4142136-0.70710678
0.51.1180340.4472136
11.41421360.70710678
22.2360680.89442719
33.16227770.9486833

Questions

What is the derivative of sqrt(x2x^{2} + 1)?

The derivative of x2+1\sqrt{x^{2} + 1} is 2x2x2+1\frac{2 x}{2 \sqrt{x^{2} + 1}}.

What is the second derivative of sqrt(x2x^{2} + 1)?

(4sqrt(x2x^{2} + 1) - 2x * 2 * 2x / (2sqrt(x2x^{2} + 1))) / (2sqrt(x2x^{2} + 1))^2.

Which rule do you use for sqrt(x2x^{2} + 1)?

The chain rule.

What is the slope of sqrt(x2x^{2} + 1) at x = 1?

0.70710678.

Related derivatives

    Derivative of sqrt(x^2 + 1): 2x / (2sqrt(x^2 + 1)) (with steps) | WordGPT