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Derivative of x51x\frac{x^{5}}{\frac{1}{x}}

The derivative of x51x\frac{x^{5}}{\frac{1}{x}} is 5x41xāˆ’x5āˆ’1x21x2\frac{5 x^{4} \frac{1}{x} - x^{5} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}.

f(x) =
ddx[x51x]=5x41xāˆ’x5āˆ’1x21x2\frac{d}{dx}\left[\frac{x^{5}}{\frac{1}{x}}\right] = \frac{5 x^{4} \frac{1}{x} - x^{5} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}

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Graph of f(x) = x^5 / (1 / x) and its derivative-3-2-1123-600-400-200200400600800slope 6f(x)f'(x)tangent at x = 1
Graph of f(x) = x^5 / (1 / x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 6.

How to work it out

  1. 1

    We differentiate f(x)=x51xf(x) = \frac{x^{5}}{\frac{1}{x}} with respect to x. The rule that applies is the quotient rule.

    f(x)=x51xf(x) = \frac{x^{5}}{\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=x5u = x^{5} and v=1xv = \frac{1}{x}.

    u=x5,v=1xu = x^{5},\quad v = \frac{1}{x}
  3. 3

    Differentiate each piece:

    uā€²=5x4,vā€²=āˆ’1x2u' = 5 x^{4},\quad v' = \frac{-1}{x^{2}}
  4. 4

    Put them in the rule:

    fā€²(x)=(5x4)(1x)āˆ’(x5)(āˆ’1x2)(1x)2f'(x) = \frac{\left(5 x^{4}\right)\left(\frac{1}{x}\right) - \left(x^{5}\right)\left(\frac{-1}{x^{2}}\right)}{\left(\frac{1}{x}\right)^{2}}
  5. 5

    Simplify:

    fā€²(x)=5x41xāˆ’x5āˆ’1x21x2f'(x) = \frac{5 x^{4} \frac{1}{x} - x^{5} \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)=(20x31x+5x4āˆ’1x2āˆ’(5x4āˆ’1x2+x5āˆ’((āˆ’2)x)(x2)2))1x2āˆ’(5x41xāˆ’x5āˆ’1x2)21xāˆ’1x2(1x2)2f''(x) = \frac{\left(20 x^{3} \frac{1}{x} + 5 x^{4} \frac{-1}{x^{2}} - \left(5 x^{4} \frac{-1}{x^{2}} + x^{5} \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}}\right)\right) \frac{1}{x}^{2} - \left(5 x^{4} \frac{1}{x} - x^{5} \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
\[ \frac{d}{dx}\left[\frac{x^{5}}{\frac{1}{x}}\right] = \frac{5 x^{4} \frac{1}{x} - x^{5} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}} \]
\[ f(x) = \frac{x^{5}}{\frac{1}{x}} \]
\[ u = x^{5},\quad v = \frac{1}{x} \]
\[ u' = 5 x^{4},\quad v' = \frac{-1}{x^{2}} \]
\[ f'(x) = \frac{\left(5 x^{4}\right)\left(\frac{1}{x}\right) - \left(x^{5}\right)\left(\frac{-1}{x^{2}}\right)}{\left(\frac{1}{x}\right)^{2}} \]
\[ f'(x) = \frac{5 x^{4} \frac{1}{x} - x^{5} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}} \]
\[ f''(x) = \frac{\left(20 x^{3} \frac{1}{x} + 5 x^{4} \frac{-1}{x^{2}} - \left(5 x^{4} \frac{-1}{x^{2}} + x^{5} \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}}\right)\right) \frac{1}{x}^{2} - \left(5 x^{4} \frac{1}{x} - x^{5} \frac{-1}{x^{2}}\right) 2 \frac{1}{x} \frac{-1}{x^{2}}}{\left(\frac{1}{x}^{2}\right)^{2}} \]

Values of f and f' for x5x^{5} / (1 / x)

xf(x)f'(x)
-264-192
-11-6
0.50.0156250.1875
116
264192
37291458

Questions

What is the derivative of x5x^{5} / (1 / x)?

The derivative of x51x\frac{x^{5}}{\frac{1}{x}} is 5x41xāˆ’x5āˆ’1x21x2\frac{5 x^{4} \frac{1}{x} - x^{5} \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}.

What is the second derivative of x5x^{5} / (1 / x)?

((20 * x3x^{3} * 1 / x + 5 * x4x^{4} * -1 / x2x^{2} - (5 * x4x^{4} * -1 / x2x^{2} + x5x^{5} * -(-2x) / (x2x^{2})^2)) * (1 / x)^2 - (5 * x4x^{4} * 1 / x - x5x^{5} * -1 / x2x^{2}) * 2 * 1 / x * -1 / x2x^{2}) / ((1 / x)^2)^2.

Which rule do you use for x5x^{5} / (1 / x)?

The quotient rule.

What is the slope of x5x^{5} / (1 / x) at x = 1?

6.

Related derivatives

    Derivative of x^5 / (1 / x): (5 * x^4 * 1 / x - x^5 * -1 / x^2) / (1 / x)^2 (with steps) | WordGPT