WordGPT

Derivative of xarctan⁡(x)\sqrt{x} \arctan\left(x\right)

The derivative of xarctan⁡(x)\sqrt{x} \arctan\left(x\right) is arctan⁡(x)2x+x11+x2\frac{\arctan\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{1 + x^{2}}.

f(x) =
ddx[xarctan⁡(x)]=arctan⁡(x)2x+x11+x2\frac{d}{dx}\left[\sqrt{x} \arctan\left(x\right)\right] = \frac{\arctan\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{1 + x^{2}}

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) = sqrt(x) * atan(x) and its derivative-3-2-1123-3-2-112slope 0.8927f(x)f'(x)tangent at x = 1
Graph of f(x) = sqrt(x) * atan(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 0.8927.

How to work it out

  1. 1

    We differentiate f(x)=xarctan⁡(x)f(x) = \sqrt{x} \arctan\left(x\right) with respect to x. The rule that applies is the product rule.

    f(x)=xarctan⁡(x)f(x) = \sqrt{x} \arctan\left(x\right)
  2. 2

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. Let u=xu = \sqrt{x} and v=arctan⁡(x)v = \arctan\left(x\right).

    u=x,v=arctan⁡(x)u = \sqrt{x},\quad v = \arctan\left(x\right)
  3. 3

    Differentiate each piece:

    u′=12x,v′=11+x2u' = \frac{1}{2 \sqrt{x}},\quad v' = \frac{1}{1 + x^{2}}
  4. 4

    Put them in the rule:

    f′(x)=(12x)(arctan⁡(x))+(x)(11+x2)f'(x) = \left(\frac{1}{2 \sqrt{x}}\right)\left(\arctan\left(x\right)\right) + \left(\sqrt{x}\right)\left(\frac{1}{1 + x^{2}}\right)
  5. 5

    Simplify:

    f′(x)=arctan⁡(x)2x+x11+x2f'(x) = \frac{\arctan\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{1 + x^{2}}
  6. 6

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

    f′′(x)=2x1+x2−arctan⁡(x)212x(2x)2+11+x22x+x−(2x)(1+x2)2f''(x) = \frac{\frac{2 \sqrt{x}}{1 + x^{2}} - \arctan\left(x\right) 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} + \frac{\frac{1}{1 + x^{2}}}{2 \sqrt{x}} + \sqrt{x} \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\sqrt{x} \arctan\left(x\right)\right] = \frac{\arctan\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{1 + x^{2}} \]
\[ f(x) = \sqrt{x} \arctan\left(x\right) \]
\[ u = \sqrt{x},\quad v = \arctan\left(x\right) \]
\[ u' = \frac{1}{2 \sqrt{x}},\quad v' = \frac{1}{1 + x^{2}} \]
\[ f'(x) = \left(\frac{1}{2 \sqrt{x}}\right)\left(\arctan\left(x\right)\right) + \left(\sqrt{x}\right)\left(\frac{1}{1 + x^{2}}\right) \]
\[ f'(x) = \frac{\arctan\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{1 + x^{2}} \]
\[ f''(x) = \frac{\frac{2 \sqrt{x}}{1 + x^{2}} - \arctan\left(x\right) 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} + \frac{\frac{1}{1 + x^{2}}}{2 \sqrt{x}} + \sqrt{x} \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}} \]

Values of f and f' for sqrt(x) * atan(x)

xf(x)f'(x)
0.50.327848370.89353379
10.785398160.89269908
21.56574470.6742789
32.16341070.53377354

Questions

What is the derivative of sqrt(x) * atan(x)?

The derivative of xarctan⁡(x)\sqrt{x} \arctan\left(x\right) is arctan⁡(x)2x+x11+x2\frac{\arctan\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{1 + x^{2}}.

What is the second derivative of sqrt(x) * atan(x)?

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

Which rule do you use for sqrt(x) * atan(x)?

The product rule.

What is the slope of sqrt(x) * atan(x) at x = 1?

0.89269908.

Related derivatives

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