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Derivative of x41x\frac{x^{4}}{\frac{1}{x}}

The derivative of x41x\frac{x^{4}}{\frac{1}{x}} is 4x31xāˆ’x4āˆ’1x21x2\frac{4 x^{3} \frac{1}{x} - x^{4} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}.

f(x) =
ddx[x41x]=4x31xāˆ’x4āˆ’1x21x2\frac{d}{dx}\left[\frac{x^{4}}{\frac{1}{x}}\right] = \frac{4 x^{3} \frac{1}{x} - x^{4} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}

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Graph of f(x) = x^4 / (1 / x) and its derivative-3-2-1123-100100200300slope 5f(x)f'(x)tangent at x = 1
Graph of f(x) = x^4 / (1 / x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 5.

How to work it out

  1. 1

    We differentiate f(x)=x41xf(x) = \frac{x^{4}}{\frac{1}{x}} with respect to x. The rule that applies is the quotient rule.

    f(x)=x41xf(x) = \frac{x^{4}}{\frac{1}{x}}
  2. 2

    Quotient rule: (uv)ā€²=uā€²vāˆ’uvā€²v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}. Let u=x4u = x^{4} and v=1xv = \frac{1}{x}.

    u=x4,v=1xu = x^{4},\quad v = \frac{1}{x}
  3. 3

    Differentiate each piece:

    uā€²=4x3,vā€²=āˆ’1x2u' = 4 x^{3},\quad v' = \frac{-1}{x^{2}}
  4. 4

    Put them in the rule:

    fā€²(x)=(4x3)(1x)āˆ’(x4)(āˆ’1x2)(1x)2f'(x) = \frac{\left(4 x^{3}\right)\left(\frac{1}{x}\right) - \left(x^{4}\right)\left(\frac{-1}{x^{2}}\right)}{\left(\frac{1}{x}\right)^{2}}
  5. 5

    Simplify:

    fā€²(x)=4x31xāˆ’x4āˆ’1x21x2f'(x) = \frac{4 x^{3} \frac{1}{x} - x^{4} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}
  6. 6

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

    fā€²ā€²(x)=(12x21x+4x3āˆ’1x2āˆ’(4x3āˆ’1x2+x4āˆ’((āˆ’2)x)(x2)2))1x2āˆ’(4x31xāˆ’x4āˆ’1x2)21xāˆ’1x2(1x2)2f''(x) = \frac{\left(12 x^{2} \frac{1}{x} + 4 x^{3} \frac{-1}{x^{2}} - \left(4 x^{3} \frac{-1}{x^{2}} + x^{4} \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}}\right)\right) \frac{1}{x}^{2} - \left(4 x^{3} \frac{1}{x} - x^{4} \frac{-1}{x^{2}}\right) 2 \frac{1}{x} \frac{-1}{x^{2}}}{\left(\frac{1}{x}^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{x^{4}}{\frac{1}{x}}\right] = \frac{4 x^{3} \frac{1}{x} - x^{4} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}} \]
\[ f(x) = \frac{x^{4}}{\frac{1}{x}} \]
\[ u = x^{4},\quad v = \frac{1}{x} \]
\[ u' = 4 x^{3},\quad v' = \frac{-1}{x^{2}} \]
\[ f'(x) = \frac{\left(4 x^{3}\right)\left(\frac{1}{x}\right) - \left(x^{4}\right)\left(\frac{-1}{x^{2}}\right)}{\left(\frac{1}{x}\right)^{2}} \]
\[ f'(x) = \frac{4 x^{3} \frac{1}{x} - x^{4} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}} \]
\[ f''(x) = \frac{\left(12 x^{2} \frac{1}{x} + 4 x^{3} \frac{-1}{x^{2}} - \left(4 x^{3} \frac{-1}{x^{2}} + x^{4} \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}}\right)\right) \frac{1}{x}^{2} - \left(4 x^{3} \frac{1}{x} - x^{4} \frac{-1}{x^{2}}\right) 2 \frac{1}{x} \frac{-1}{x^{2}}}{\left(\frac{1}{x}^{2}\right)^{2}} \]

Values of f and f' for x4x^{4} / (1 / x)

xf(x)f'(x)
-2-3280
-1-15
0.50.031250.3125
115
23280
3243405

Questions

What is the derivative of x4x^{4} / (1 / x)?

The derivative of x41x\frac{x^{4}}{\frac{1}{x}} is 4x31xāˆ’x4āˆ’1x21x2\frac{4 x^{3} \frac{1}{x} - x^{4} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}.

What is the second derivative of x4x^{4} / (1 / x)?

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

Which rule do you use for x4x^{4} / (1 / x)?

The quotient rule.

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

5.

Related derivatives

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