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Derivative of sin⁡(x)tan⁡(x)\sin\left(x\right) \tan\left(x\right)

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

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

Type any function of x; the derivative is computed symbolically in your browser.

Graph of f(x) = sin(x) * tan(x) and its derivative-3-2-1123-20-101020slope 3.724f(x)f'(x)tangent at x = 1
Graph of f(x) = sin(x) * tan(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 3.7239.

How to work it out

  1. 1

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

    f(x)=sin⁡(x)tan⁡(x)f(x) = \sin\left(x\right) \tan\left(x\right)
  2. 2

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. Let u=sin⁡(x)u = \sin\left(x\right) and v=tan⁡(x)v = \tan\left(x\right).

    u=sin⁡(x),v=tan⁡(x)u = \sin\left(x\right),\quad v = \tan\left(x\right)
  3. 3

    Differentiate each piece:

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

    Put them in the rule:

    f′(x)=(cos⁡(x))(tan⁡(x))+(sin⁡(x))(1cos⁡(x)2)f'(x) = \left(\cos\left(x\right)\right)\left(\tan\left(x\right)\right) + \left(\sin\left(x\right)\right)\left(\frac{1}{\cos\left(x\right)^{2}}\right)
  5. 5

    Simplify:

    f′(x)=cos⁡(x)tan⁡(x)+sin⁡(x)1cos⁡(x)2f'(x) = \cos\left(x\right) \tan\left(x\right) + \sin\left(x\right) \frac{1}{\cos\left(x\right)^{2}}
  6. 6

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

    f′′(x)=−(sin⁡(x)tan⁡(x))+cos⁡(x)1cos⁡(x)2+cos⁡(x)1cos⁡(x)2+sin⁡(x)−(2cos⁡(x)−sin⁡(x))(cos⁡(x)2)2f''(x) = -\left(\sin\left(x\right) \tan\left(x\right)\right) + \cos\left(x\right) \frac{1}{\cos\left(x\right)^{2}} + \cos\left(x\right) \frac{1}{\cos\left(x\right)^{2}} + \sin\left(x\right) \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
\[ \frac{d}{dx}\left[\sin\left(x\right) \tan\left(x\right)\right] = \cos\left(x\right) \tan\left(x\right) + \sin\left(x\right) \frac{1}{\cos\left(x\right)^{2}} \]
\[ f(x) = \sin\left(x\right) \tan\left(x\right) \]
\[ u = \sin\left(x\right),\quad v = \tan\left(x\right) \]
\[ u' = \cos\left(x\right),\quad v' = \frac{1}{\cos\left(x\right)^{2}} \]
\[ f'(x) = \left(\cos\left(x\right)\right)\left(\tan\left(x\right)\right) + \left(\sin\left(x\right)\right)\left(\frac{1}{\cos\left(x\right)^{2}}\right) \]
\[ f'(x) = \cos\left(x\right) \tan\left(x\right) + \sin\left(x\right) \frac{1}{\cos\left(x\right)^{2}} \]
\[ f''(x) = -\left(\sin\left(x\right) \tan\left(x\right)\right) + \cos\left(x\right) \frac{1}{\cos\left(x\right)^{2}} + \cos\left(x\right) \frac{1}{\cos\left(x\right)^{2}} + \sin\left(x\right) \frac{-\left(2 \cos\left(x\right) -\sin\left(x\right)\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
-2-1.9868511-6.1599438
-11.3105134-3.7239457
0.50.261911371.1019339
11.31051343.7239457
2-1.98685116.1599438
3-0.0201161690.28510751

Questions

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

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

What is the second derivative of sin(x) * tan(x)?

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

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

The product rule.

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

3.7239457.

Related derivatives

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