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Derivative of xtan⁡(x)\frac{\sqrt{x}}{\tan\left(x\right)}

The derivative of xtan⁡(x)\frac{\sqrt{x}}{\tan\left(x\right)} is tan⁡(x)2x−x1cos⁡(x)2tan⁡(x)2\frac{\frac{\tan\left(x\right)}{2 \sqrt{x}} - \sqrt{x} \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}}.

f(x) =
ddx[xtan⁡(x)]=tan⁡(x)2x−x1cos⁡(x)2tan⁡(x)2\frac{d}{dx}\left[\frac{\sqrt{x}}{\tan\left(x\right)}\right] = \frac{\frac{\tan\left(x\right)}{2 \sqrt{x}} - \sqrt{x} \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}}

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Graph of f(x) = sqrt(x) / tan(x) and its derivative-3-2-1123-10-55slope -1.091f(x)f'(x)tangent at x = 1
Graph of f(x) = sqrt(x) / tan(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -1.0912.

How to work it out

  1. 1

    We differentiate f(x)=xtan⁡(x)f(x) = \frac{\sqrt{x}}{\tan\left(x\right)} with respect to x. The rule that applies is the quotient rule.

    f(x)=xtan⁡(x)f(x) = \frac{\sqrt{x}}{\tan\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=xu = \sqrt{x} and v=tan⁡(x)v = \tan\left(x\right).

    u=x,v=tan⁡(x)u = \sqrt{x},\quad v = \tan\left(x\right)
  3. 3

    Differentiate each piece:

    u′=12x,v′=1cos⁡(x)2u' = \frac{1}{2 \sqrt{x}},\quad v' = \frac{1}{\cos\left(x\right)^{2}}
  4. 4

    Put them in the rule:

    f′(x)=(12x)(tan⁡(x))−(x)(1cos⁡(x)2)(tan⁡(x))2f'(x) = \frac{\left(\frac{1}{2 \sqrt{x}}\right)\left(\tan\left(x\right)\right) - \left(\sqrt{x}\right)\left(\frac{1}{\cos\left(x\right)^{2}}\right)}{\left(\tan\left(x\right)\right)^{2}}
  5. 5

    Simplify:

    f′(x)=tan⁡(x)2x−x1cos⁡(x)2tan⁡(x)2f'(x) = \frac{\frac{\tan\left(x\right)}{2 \sqrt{x}} - \sqrt{x} \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}}
  6. 6

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

    f′′(x)=(2xcos⁡(x)2−tan⁡(x)212x(2x)2−(1cos⁡(x)22x+x−(2cos⁡(x)−sin⁡(x))(cos⁡(x)2)2))tan⁡(x)2−(tan⁡(x)2x−x1cos⁡(x)2)2tan⁡(x)1cos⁡(x)2(tan⁡(x)2)2f''(x) = \frac{\left(\frac{\frac{2 \sqrt{x}}{\cos\left(x\right)^{2}} - \tan\left(x\right) 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} - \left(\frac{\frac{1}{\cos\left(x\right)^{2}}}{2 \sqrt{x}} + \sqrt{x} \frac{-\left(2 \cos\left(x\right) -\sin\left(x\right)\right)}{\left(\cos\left(x\right)^{2}\right)^{2}}\right)\right) \tan\left(x\right)^{2} - \left(\frac{\tan\left(x\right)}{2 \sqrt{x}} - \sqrt{x} \frac{1}{\cos\left(x\right)^{2}}\right) 2 \tan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\left(\tan\left(x\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\sqrt{x}}{\tan\left(x\right)}\right] = \frac{\frac{\tan\left(x\right)}{2 \sqrt{x}} - \sqrt{x} \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}} \]
\[ f(x) = \frac{\sqrt{x}}{\tan\left(x\right)} \]
\[ u = \sqrt{x},\quad v = \tan\left(x\right) \]
\[ u' = \frac{1}{2 \sqrt{x}},\quad v' = \frac{1}{\cos\left(x\right)^{2}} \]
\[ f'(x) = \frac{\left(\frac{1}{2 \sqrt{x}}\right)\left(\tan\left(x\right)\right) - \left(\sqrt{x}\right)\left(\frac{1}{\cos\left(x\right)^{2}}\right)}{\left(\tan\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{\frac{\tan\left(x\right)}{2 \sqrt{x}} - \sqrt{x} \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{\frac{2 \sqrt{x}}{\cos\left(x\right)^{2}} - \tan\left(x\right) 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} - \left(\frac{\frac{1}{\cos\left(x\right)^{2}}}{2 \sqrt{x}} + \sqrt{x} \frac{-\left(2 \cos\left(x\right) -\sin\left(x\right)\right)}{\left(\cos\left(x\right)^{2}\right)^{2}}\right)\right) \tan\left(x\right)^{2} - \left(\frac{\tan\left(x\right)}{2 \sqrt{x}} - \sqrt{x} \frac{1}{\cos\left(x\right)^{2}}\right) 2 \tan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\left(\tan\left(x\right)^{2}\right)^{2}} \]

Values of f and f' for sqrt(x) / tan(x)

xf(x)f'(x)
0.51.2943503-1.7820488
10.64209262-1.0912366
2-0.64722552-1.8722276
3-12.150774-88.997927

Questions

What is the derivative of sqrt(x) / tan(x)?

The derivative of xtan⁡(x)\frac{\sqrt{x}}{\tan\left(x\right)} is tan⁡(x)2x−x1cos⁡(x)2tan⁡(x)2\frac{\frac{\tan\left(x\right)}{2 \sqrt{x}} - \sqrt{x} \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}}.

What is the second derivative of sqrt(x) / tan(x)?

(((2sqrt(x) / cos(x)^2 - tan(x) * 2 * 1 / (2sqrt(x))) / (2sqrt(x))^2 - (1 / cos(x)^2 / (2sqrt(x)) + sqrt(x) * -(2cos(x) * -sin(x)) / (cos(x)^2)^2)) * tan(x)^2 - (tan(x) / (2sqrt(x)) - sqrt(x) * 1 / cos(x)^2) * 2tan(x) * 1 / cos(x)^2) / (tan(x)^2)^2.

Which rule do you use for sqrt(x) / tan(x)?

The quotient rule.

What is the slope of sqrt(x) / tan(x) at x = 1?

-1.0912366.

Related derivatives

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