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

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

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

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Graph of f(x) = sqrt(x) * tan(x) and its derivative-3-2-1123-20-101020304050slope 4.204f(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) = 4.2042.

How to work it out

  1. 1

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

    f(x)=xtan⁡(x)f(x) = \sqrt{x} \tan\left(x\right)
  2. 2

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. 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)f'(x) = \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)
  5. 5

    Simplify:

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

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

xf(x)f'(x)
0.50.38629421.3044345
11.55740774.2042227
2-3.0901137.3937054
3-0.246897861.7260956

Questions

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

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

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

The product rule.

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

4.2042227.

Related derivatives

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