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Derivative of x31x\frac{x^{3}}{\frac{1}{x}}

The derivative of x31x\frac{x^{3}}{\frac{1}{x}} is 3x21xāˆ’x3āˆ’1x21x2\frac{3 x^{2} \frac{1}{x} - x^{3} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}.

f(x) =
ddx[x31x]=3x21xāˆ’x3āˆ’1x21x2\frac{d}{dx}\left[\frac{x^{3}}{\frac{1}{x}}\right] = \frac{3 x^{2} \frac{1}{x} - x^{3} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}

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Graph of f(x) = x^3 / (1 / x) and its derivative-3-2-1123-60-40-2020406080slope 4f(x)f'(x)tangent at x = 1
Graph of f(x) = x^3 / (1 / x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 4.

How to work it out

  1. 1

    We differentiate f(x)=x31xf(x) = \frac{x^{3}}{\frac{1}{x}} with respect to x. The rule that applies is the quotient rule.

    f(x)=x31xf(x) = \frac{x^{3}}{\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=x3u = x^{3} and v=1xv = \frac{1}{x}.

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

    Differentiate each piece:

    uā€²=3x2,vā€²=āˆ’1x2u' = 3 x^{2},\quad v' = \frac{-1}{x^{2}}
  4. 4

    Put them in the rule:

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

    Simplify:

    fā€²(x)=3x21xāˆ’x3āˆ’1x21x2f'(x) = \frac{3 x^{2} \frac{1}{x} - x^{3} \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)=(6x1x+3x2āˆ’1x2āˆ’(3x2āˆ’1x2+x3āˆ’((āˆ’2)x)(x2)2))1x2āˆ’(3x21xāˆ’x3āˆ’1x2)21xāˆ’1x2(1x2)2f''(x) = \frac{\left(6 x \frac{1}{x} + 3 x^{2} \frac{-1}{x^{2}} - \left(3 x^{2} \frac{-1}{x^{2}} + x^{3} \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}}\right)\right) \frac{1}{x}^{2} - \left(3 x^{2} \frac{1}{x} - x^{3} \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
\[ \frac{d}{dx}\left[\frac{x^{3}}{\frac{1}{x}}\right] = \frac{3 x^{2} \frac{1}{x} - x^{3} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}} \]
\[ f(x) = \frac{x^{3}}{\frac{1}{x}} \]
\[ u = x^{3},\quad v = \frac{1}{x} \]
\[ u' = 3 x^{2},\quad v' = \frac{-1}{x^{2}} \]
\[ f'(x) = \frac{\left(3 x^{2}\right)\left(\frac{1}{x}\right) - \left(x^{3}\right)\left(\frac{-1}{x^{2}}\right)}{\left(\frac{1}{x}\right)^{2}} \]
\[ f'(x) = \frac{3 x^{2} \frac{1}{x} - x^{3} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}} \]
\[ f''(x) = \frac{\left(6 x \frac{1}{x} + 3 x^{2} \frac{-1}{x^{2}} - \left(3 x^{2} \frac{-1}{x^{2}} + x^{3} \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}}\right)\right) \frac{1}{x}^{2} - \left(3 x^{2} \frac{1}{x} - x^{3} \frac{-1}{x^{2}}\right) 2 \frac{1}{x} \frac{-1}{x^{2}}}{\left(\frac{1}{x}^{2}\right)^{2}} \]

Values of f and f' for x3x^{3} / (1 / x)

xf(x)f'(x)
-216-32
-11-4
0.50.06250.5
114
21632
381108

Questions

What is the derivative of x3x^{3} / (1 / x)?

The derivative of x31x\frac{x^{3}}{\frac{1}{x}} is 3x21xāˆ’x3āˆ’1x21x2\frac{3 x^{2} \frac{1}{x} - x^{3} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}.

What is the second derivative of x3x^{3} / (1 / x)?

((6x * 1 / x + 3 * x2x^{2} * -1 / x2x^{2} - (3 * x2x^{2} * -1 / x2x^{2} + x3x^{3} * -(-2x) / (x2x^{2})^2)) * (1 / x)^2 - (3 * x2x^{2} * 1 / x - x3x^{3} * -1 / x2x^{2}) * 2 * 1 / x * -1 / x2x^{2}) / ((1 / x)^2)^2.

Which rule do you use for x3x^{3} / (1 / x)?

The quotient rule.

What is the slope of x3x^{3} / (1 / x) at x = 1?

4.

Related derivatives

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