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Derivative of x2cos⁡(x)\frac{x^{2}}{\cos\left(x\right)}

The derivative of x2cos⁡(x)\frac{x^{2}}{\cos\left(x\right)} is 2xcos⁡(x)−x2−sin⁡(x)cos⁡(x)2\frac{2 x \cos\left(x\right) - x^{2} -\sin\left(x\right)}{\cos\left(x\right)^{2}}.

f(x) =
ddx[x2cos⁡(x)]=2xcos⁡(x)−x2−sin⁡(x)cos⁡(x)2\frac{d}{dx}\left[\frac{x^{2}}{\cos\left(x\right)}\right] = \frac{2 x \cos\left(x\right) - x^{2} -\sin\left(x\right)}{\cos\left(x\right)^{2}}

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Graph of f(x) = x^2 / cos(x) and its derivative-3-2-1123-60-40-20204060slope 6.584f(x)f'(x)tangent at x = 1
Graph of f(x) = x^2 / cos(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 6.5841.

How to work it out

  1. 1

    We differentiate f(x)=x2cos⁡(x)f(x) = \frac{x^{2}}{\cos\left(x\right)} with respect to x. The rule that applies is the quotient rule.

    f(x)=x2cos⁡(x)f(x) = \frac{x^{2}}{\cos\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=x2u = x^{2} and v=cos⁡(x)v = \cos\left(x\right).

    u=x2,v=cos⁡(x)u = x^{2},\quad v = \cos\left(x\right)
  3. 3

    Differentiate each piece:

    u′=2x,v′=−sin⁡(x)u' = 2 x,\quad v' = -\sin\left(x\right)
  4. 4

    Put them in the rule:

    f′(x)=(2x)(cos⁡(x))−(x2)(−sin⁡(x))(cos⁡(x))2f'(x) = \frac{\left(2 x\right)\left(\cos\left(x\right)\right) - \left(x^{2}\right)\left(-\sin\left(x\right)\right)}{\left(\cos\left(x\right)\right)^{2}}
  5. 5

    Simplify:

    f′(x)=2xcos⁡(x)−x2−sin⁡(x)cos⁡(x)2f'(x) = \frac{2 x \cos\left(x\right) - x^{2} -\sin\left(x\right)}{\cos\left(x\right)^{2}}
  6. 6

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

    f′′(x)=(2cos⁡(x)+2x−sin⁡(x)−(2x−sin⁡(x)+x2−cos⁡(x)))cos⁡(x)2−(2xcos⁡(x)−x2−sin⁡(x))2cos⁡(x)−sin⁡(x)(cos⁡(x)2)2f''(x) = \frac{\left(2 \cos\left(x\right) + 2 x -\sin\left(x\right) - \left(2 x -\sin\left(x\right) + x^{2} -\cos\left(x\right)\right)\right) \cos\left(x\right)^{2} - \left(2 x \cos\left(x\right) - x^{2} -\sin\left(x\right)\right) 2 \cos\left(x\right) -\sin\left(x\right)}{\left(\cos\left(x\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{x^{2}}{\cos\left(x\right)}\right] = \frac{2 x \cos\left(x\right) - x^{2} -\sin\left(x\right)}{\cos\left(x\right)^{2}} \]
\[ f(x) = \frac{x^{2}}{\cos\left(x\right)} \]
\[ u = x^{2},\quad v = \cos\left(x\right) \]
\[ u' = 2 x,\quad v' = -\sin\left(x\right) \]
\[ f'(x) = \frac{\left(2 x\right)\left(\cos\left(x\right)\right) - \left(x^{2}\right)\left(-\sin\left(x\right)\right)}{\left(\cos\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{2 x \cos\left(x\right) - x^{2} -\sin\left(x\right)}{\cos\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(2 \cos\left(x\right) + 2 x -\sin\left(x\right) - \left(2 x -\sin\left(x\right) + x^{2} -\cos\left(x\right)\right)\right) \cos\left(x\right)^{2} - \left(2 x \cos\left(x\right) - x^{2} -\sin\left(x\right)\right) 2 \cos\left(x\right) -\sin\left(x\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} \]

Values of f and f' for x2x^{2} / cos(x)

xf(x)f'(x)
-2-9.6119918-11.390594
-11.8508157-6.5841061
0.50.284873481.295121
11.85081576.5841061
2-9.611991811.390594
3-9.090978-4.7647645

Questions

What is the derivative of x2x^{2} / cos(x)?

The derivative of x2cos⁡(x)\frac{x^{2}}{\cos\left(x\right)} is 2xcos⁡(x)−x2−sin⁡(x)cos⁡(x)2\frac{2 x \cos\left(x\right) - x^{2} -\sin\left(x\right)}{\cos\left(x\right)^{2}}.

What is the second derivative of x2x^{2} / cos(x)?

((2cos(x) + 2x * -sin(x) - (2x * -sin(x) + x2x^{2} * -cos(x))) * cos(x)^2 - (2x * cos(x) - x2x^{2} * -sin(x)) * 2cos(x) * -sin(x)) / (cos(x)^2)^2.

Which rule do you use for x2x^{2} / cos(x)?

The quotient rule.

What is the slope of x2x^{2} / cos(x) at x = 1?

6.5841061.

Related derivatives

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