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

The derivative of x4ln⁡(x)x^{4} \ln\left(x\right) is 4x3ln⁡(x)+x41x4 x^{3} \ln\left(x\right) + x^{4} \frac{1}{x}.

f(x) =
ddx[x4ln⁡(x)]=4x3ln⁡(x)+x41x\frac{d}{dx}\left[x^{4} \ln\left(x\right)\right] = 4 x^{3} \ln\left(x\right) + x^{4} \frac{1}{x}

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Graph of f(x) = x^4 * ln(x) and its derivative-3-2-112320406080100slope 1f(x)f'(x)tangent at x = 1
Graph of f(x) = x^4 * 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)=x4ln⁡(x)f(x) = x^{4} \ln\left(x\right) with respect to x. The rule that applies is the product rule.

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

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. 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)f'(x) = \left(4 x^{3}\right)\left(\ln\left(x\right)\right) + \left(x^{4}\right)\left(\frac{1}{x}\right)
  5. 5

    Simplify:

    f′(x)=4x3ln⁡(x)+x41xf'(x) = 4 x^{3} \ln\left(x\right) + x^{4} \frac{1}{x}
  6. 6

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

    f′′(x)=12x2ln⁡(x)+4x31x+4x31x+x4−1x2f''(x) = 12 x^{2} \ln\left(x\right) + 4 x^{3} \frac{1}{x} + 4 x^{3} \frac{1}{x} + x^{4} \frac{-1}{x^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[x^{4} \ln\left(x\right)\right] = 4 x^{3} \ln\left(x\right) + x^{4} \frac{1}{x} \]
\[ f(x) = x^{4} \ln\left(x\right) \]
\[ u = x^{4},\quad v = \ln\left(x\right) \]
\[ u' = 4 x^{3},\quad v' = \frac{1}{x} \]
\[ f'(x) = \left(4 x^{3}\right)\left(\ln\left(x\right)\right) + \left(x^{4}\right)\left(\frac{1}{x}\right) \]
\[ f'(x) = 4 x^{3} \ln\left(x\right) + x^{4} \frac{1}{x} \]
\[ f''(x) = 12 x^{2} \ln\left(x\right) + 4 x^{3} \frac{1}{x} + 4 x^{3} \frac{1}{x} + x^{4} \frac{-1}{x^{2}} \]

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

xf(x)f'(x)
0.5-0.043321699-0.22157359
101
211.09035530.18071
388.987595145.65013

Questions

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

The derivative of x4ln⁡(x)x^{4} \ln\left(x\right) is 4x3ln⁡(x)+x41x4 x^{3} \ln\left(x\right) + x^{4} \frac{1}{x}.

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}.

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

The product rule.

What is the slope of x4x^{4} * ln(x) at x = 1?

1.

Related derivatives

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