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Derivative of xln⁡(x)\sqrt{x} \ln\left(x\right)

The derivative of xln⁡(x)\sqrt{x} \ln\left(x\right) is ln⁡(x)2x+x1x\frac{\ln\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{x}.

f(x) =
ddx[xln⁡(x)]=ln⁡(x)2x+x1x\frac{d}{dx}\left[\sqrt{x} \ln\left(x\right)\right] = \frac{\ln\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{x}

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Graph of f(x) = sqrt(x) * ln(x) and its derivative-3-2-1123-4-3-2-112slope 1f(x)f'(x)tangent at x = 1
Graph of f(x) = sqrt(x) * ln(x) 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)=xln⁡(x)f(x) = \sqrt{x} \ln\left(x\right) with respect to x. The rule that applies is the product rule.

    f(x)=xln⁡(x)f(x) = \sqrt{x} \ln\left(x\right)
  2. 2

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

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

    Differentiate each piece:

    u′=12x,v′=1xu' = \frac{1}{2 \sqrt{x}},\quad v' = \frac{1}{x}
  4. 4

    Put them in the rule:

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

    Simplify:

    f′(x)=ln⁡(x)2x+x1xf'(x) = \frac{\ln\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{x}
  6. 6

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

    f′′(x)=2xx−ln⁡(x)212x(2x)2+1x2x+x−1x2f''(x) = \frac{\frac{2 \sqrt{x}}{x} - \ln\left(x\right) 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} + \frac{\frac{1}{x}}{2 \sqrt{x}} + \sqrt{x} \frac{-1}{x^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\sqrt{x} \ln\left(x\right)\right] = \frac{\ln\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{x} \]
\[ f(x) = \sqrt{x} \ln\left(x\right) \]
\[ u = \sqrt{x},\quad v = \ln\left(x\right) \]
\[ u' = \frac{1}{2 \sqrt{x}},\quad v' = \frac{1}{x} \]
\[ f'(x) = \left(\frac{1}{2 \sqrt{x}}\right)\left(\ln\left(x\right)\right) + \left(\sqrt{x}\right)\left(\frac{1}{x}\right) \]
\[ f'(x) = \frac{\ln\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{x} \]
\[ f''(x) = \frac{\frac{2 \sqrt{x}}{x} - \ln\left(x\right) 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} + \frac{\frac{1}{x}}{2 \sqrt{x}} + \sqrt{x} \frac{-1}{x^{2}} \]

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

xf(x)f'(x)
0.5-0.490129070.92408449
101
20.980258140.95217132
31.90285230.89449232

Questions

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

The derivative of xln⁡(x)\sqrt{x} \ln\left(x\right) is ln⁡(x)2x+x1x\frac{\ln\left(x\right)}{2 \sqrt{x}} + \sqrt{x} \frac{1}{x}.

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

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

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

The product rule.

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

1.

Related derivatives

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