WordGPT

Derivative of ln⁡(x2+1)\ln\left(x^{2} + 1\right)

The derivative of ln⁡(x2+1)\ln\left(x^{2} + 1\right) is 2xx2+1\frac{2 x}{x^{2} + 1}.

f(x) =
ddx[ln⁡(x2+1)]=2xx2+1\frac{d}{dx}\left[\ln\left(x^{2} + 1\right)\right] = \frac{2 x}{x^{2} + 1}

Type any function of x; the derivative is computed symbolically in your browser.

Explain it step by step in chatMake a short video of it
Graph of f(x) = ln(x^2 + 1) and its derivative-3-2-1123-3-2-112slope 1f(x)f'(x)tangent at x = 1
Graph of f(x) = ln(x^2 + 1) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 1.

How to work it out

  1. 1

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

    f(x)=ln⁡(x2+1)f(x) = \ln\left(x^{2} + 1\right)
  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,g′(x)=2xg(x) = x^{2} + 1,\quad g'(x) = 2 x
  3. 3

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

    f′(x)=[1u]u=g(x)⋅g′(x)f'(x) = \left[\frac{1}{u}\right]_{u=g(x)} \cdot g'(x)
  4. 4

    Simplify:

    f′(x)=2xx2+1f'(x) = \frac{2 x}{x^{2} + 1}
  5. 5

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

    f′′(x)=2(x2+1)−2x2x(x2+1)2f''(x) = \frac{2 \left(x^{2} + 1\right) - 2 x 2 x}{\left(x^{2} + 1\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\ln\left(x^{2} + 1\right)\right] = \frac{2 x}{x^{2} + 1} \]
\[ f(x) = \ln\left(x^{2} + 1\right) \]
\[ g(x) = x^{2} + 1,\quad g'(x) = 2 x \]
\[ f'(x) = \left[\frac{1}{u}\right]_{u=g(x)} \cdot g'(x) \]
\[ f'(x) = \frac{2 x}{x^{2} + 1} \]
\[ f''(x) = \frac{2 \left(x^{2} + 1\right) - 2 x 2 x}{\left(x^{2} + 1\right)^{2}} \]

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

xf(x)f'(x)
-21.6094379-0.8
-10.69314718-1
0.50.223143550.8
10.693147181
21.60943790.8
32.30258510.6

Questions

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

The derivative of ln⁡(x2+1)\ln\left(x^{2} + 1\right) is 2xx2+1\frac{2 x}{x^{2} + 1}.

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

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

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

The chain rule.

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

1.

Related derivatives

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