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Derivative of tan⁡(x2+1)\tan\left(x^{2} + 1\right)

The derivative of tan⁡(x2+1)\tan\left(x^{2} + 1\right) is 2xcos⁡(x2+1)2\frac{2 x}{\cos\left(x^{2} + 1\right)^{2}}.

f(x) =
ddx[tan⁡(x2+1)]=2xcos⁡(x2+1)2\frac{d}{dx}\left[\tan\left(x^{2} + 1\right)\right] = \frac{2 x}{\cos\left(x^{2} + 1\right)^{2}}

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Graph of f(x) = tan(x^2 + 1) and its derivative-3-2-1123-60-40-20204060slope 11.55f(x)f'(x)tangent at x = 1
Graph of f(x) = tan(x^2 + 1) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 11.549.

How to work it out

  1. 1

    We differentiate f(x)=tan⁡(x2+1)f(x) = \tan\left(x^{2} + 1\right) with respect to x. The rule that applies is the chain rule.

    f(x)=tan⁡(x2+1)f(x) = \tan\left(x^{2} + 1\right)
  2. 2

    Chain rule: differentiate the outer function, keep the inside, then multiply by the derivative of the inside. The inside is g(x)=x2+1g(x) = x^{2} + 1.

    g(x)=x2+1,g′(x)=2xg(x) = x^{2} + 1,\quad g'(x) = 2 x
  3. 3

    The outer function is tan(u); its derivative evaluated at g(x), times g'(x):

    f′(x)=[1cos⁡(u)2]u=g(x)⋅g′(x)f'(x) = \left[\frac{1}{\cos\left(u\right)^{2}}\right]_{u=g(x)} \cdot g'(x)
  4. 4

    Simplify:

    f′(x)=2xcos⁡(x2+1)2f'(x) = \frac{2 x}{\cos\left(x^{2} + 1\right)^{2}}
  5. 5

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

    f′′(x)=2cos⁡(x2+1)2−2x2cos⁡(x2+1)−(sin⁡(x2+1)2x)(cos⁡(x2+1)2)2f''(x) = \frac{2 \cos\left(x^{2} + 1\right)^{2} - 2 x 2 \cos\left(x^{2} + 1\right) -\left(\sin\left(x^{2} + 1\right) 2 x\right)}{\left(\cos\left(x^{2} + 1\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\tan\left(x^{2} + 1\right)\right] = \frac{2 x}{\cos\left(x^{2} + 1\right)^{2}} \]
\[ f(x) = \tan\left(x^{2} + 1\right) \]
\[ g(x) = x^{2} + 1,\quad g'(x) = 2 x \]
\[ f'(x) = \left[\frac{1}{\cos\left(u\right)^{2}}\right]_{u=g(x)} \cdot g'(x) \]
\[ f'(x) = \frac{2 x}{\cos\left(x^{2} + 1\right)^{2}} \]
\[ f''(x) = \frac{2 \cos\left(x^{2} + 1\right)^{2} - 2 x 2 \cos\left(x^{2} + 1\right) -\left(\sin\left(x^{2} + 1\right) 2 x\right)}{\left(\cos\left(x^{2} + 1\right)^{2}\right)^{2}} \]

Values of f and f' for tan(x2x^{2} + 1)

xf(x)f'(x)
-2-3.380515-49.711527
-1-2.1850399-11.548798
0.53.009569710.05751
1-2.185039911.548798
2-3.38051549.711527
30.648360838.5222306

Questions

What is the derivative of tan(x2x^{2} + 1)?

The derivative of tan⁡(x2+1)\tan\left(x^{2} + 1\right) is 2xcos⁡(x2+1)2\frac{2 x}{\cos\left(x^{2} + 1\right)^{2}}.

What is the second derivative of tan(x2x^{2} + 1)?

(2 * cos(x2x^{2} + 1)^2 - 2x * 2cos(x2x^{2} + 1) * -(sin(x2x^{2} + 1) * 2x)) / (cos(x2x^{2} + 1)^2)^2.

Which rule do you use for tan(x2x^{2} + 1)?

The chain rule.

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

11.548798.

Related derivatives

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