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Derivative of x4ln⁡(x)\frac{x^{4}}{\ln\left(x\right)}

The derivative of x4ln⁡(x)\frac{x^{4}}{\ln\left(x\right)} is 4x3ln⁡(x)−x41xln⁡(x)2\frac{4 x^{3} \ln\left(x\right) - x^{4} \frac{1}{x}}{\ln\left(x\right)^{2}}.

f(x) =
ddx[x4ln⁡(x)]=4x3ln⁡(x)−x41xln⁡(x)2\frac{d}{dx}\left[\frac{x^{4}}{\ln\left(x\right)}\right] = \frac{4 x^{3} \ln\left(x\right) - x^{4} \frac{1}{x}}{\ln\left(x\right)^{2}}

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Graph of f(x) = x^4 / ln(x) and its derivative-3-2-1123-20204060slope -0.9815f(x)f'(x)tangent at x = 0.5
Graph of f(x) = x^4 / ln(x) and its derivative. The dashed line is the tangent at x = 0.5; its slope is f'(0.5) = -0.98152.

How to work it out

  1. 1

    We differentiate f(x)=x4ln⁡(x)f(x) = \frac{x^{4}}{\ln\left(x\right)} with respect to x. The rule that applies is the quotient rule.

    f(x)=x4ln⁡(x)f(x) = \frac{x^{4}}{\ln\left(x\right)}
  2. 2

    Quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}. Let u=x4u = x^{4} and v=ln⁡(x)v = \ln\left(x\right).

    u=x4,v=ln⁡(x)u = x^{4},\quad v = \ln\left(x\right)
  3. 3

    Differentiate each piece:

    u′=4x3,v′=1xu' = 4 x^{3},\quad v' = \frac{1}{x}
  4. 4

    Put them in the rule:

    f′(x)=(4x3)(ln⁡(x))−(x4)(1x)(ln⁡(x))2f'(x) = \frac{\left(4 x^{3}\right)\left(\ln\left(x\right)\right) - \left(x^{4}\right)\left(\frac{1}{x}\right)}{\left(\ln\left(x\right)\right)^{2}}
  5. 5

    Simplify:

    f′(x)=4x3ln⁡(x)−x41xln⁡(x)2f'(x) = \frac{4 x^{3} \ln\left(x\right) - x^{4} \frac{1}{x}}{\ln\left(x\right)^{2}}
  6. 6

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

    f′′(x)=(12x2ln⁡(x)+4x31x−(4x31x+x4−1x2))ln⁡(x)2−(4x3ln⁡(x)−x41x)2ln⁡(x)1x(ln⁡(x)2)2f''(x) = \frac{\left(12 x^{2} \ln\left(x\right) + 4 x^{3} \frac{1}{x} - \left(4 x^{3} \frac{1}{x} + x^{4} \frac{-1}{x^{2}}\right)\right) \ln\left(x\right)^{2} - \left(4 x^{3} \ln\left(x\right) - x^{4} \frac{1}{x}\right) 2 \ln\left(x\right) \frac{1}{x}}{\left(\ln\left(x\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{x^{4}}{\ln\left(x\right)}\right] = \frac{4 x^{3} \ln\left(x\right) - x^{4} \frac{1}{x}}{\ln\left(x\right)^{2}} \]
\[ f(x) = \frac{x^{4}}{\ln\left(x\right)} \]
\[ u = x^{4},\quad v = \ln\left(x\right) \]
\[ u' = 4 x^{3},\quad v' = \frac{1}{x} \]
\[ f'(x) = \frac{\left(4 x^{3}\right)\left(\ln\left(x\right)\right) - \left(x^{4}\right)\left(\frac{1}{x}\right)}{\left(\ln\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{4 x^{3} \ln\left(x\right) - x^{4} \frac{1}{x}}{\ln\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(12 x^{2} \ln\left(x\right) + 4 x^{3} \frac{1}{x} - \left(4 x^{3} \frac{1}{x} + x^{4} \frac{-1}{x^{2}}\right)\right) \ln\left(x\right)^{2} - \left(4 x^{3} \ln\left(x\right) - x^{4} \frac{1}{x}\right) 2 \ln\left(x\right) \frac{1}{x}}{\left(\ln\left(x\right)^{2}\right)^{2}} \]

Values of f and f' for x4x^{4} / ln(x)

xf(x)f'(x)
0.5-0.09016844-0.98151864
223.08312129.515289
373.72937775.935379

Questions

What is the derivative of x4x^{4} / ln(x)?

The derivative of x4ln⁡(x)\frac{x^{4}}{\ln\left(x\right)} is 4x3ln⁡(x)−x41xln⁡(x)2\frac{4 x^{3} \ln\left(x\right) - x^{4} \frac{1}{x}}{\ln\left(x\right)^{2}}.

What is the second derivative of x4x^{4} / ln(x)?

((12 * x2x^{2} * ln(x) + 4 * x3x^{3} * 1 / x - (4 * x3x^{3} * 1 / x + x4x^{4} * -1 / x2x^{2})) * ln(x)^2 - (4 * x3x^{3} * ln(x) - x4x^{4} * 1 / x) * 2ln(x) * 1 / x) / (ln(x)^2)^2.

Which rule do you use for x4x^{4} / ln(x)?

The quotient rule.

What is the slope of x4x^{4} / ln(x) at x = 0.5?

-0.98151864.

Related derivatives

    Derivative of x^4 / ln(x): (4 * x^3 * ln(x) - x^4 * 1 / x) / ln(x)^2 (with steps) | WordGPT