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Given f(x)=2sec(2x), find f^(')(x). Answer: f^(')(x)=

Given f(x)=2sec⁡(2x) f(x)=2 \sec (2 x) , find f′(x) f^{\prime}(x) .\newlineAnswer: f′(x)= f^{\prime}(x)=

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Evaluate a nonlinear functionright chevron icon

Full solution

Q. Given f(x)=2sec⁡(2x) f(x)=2 \sec (2 x) , find f′(x) f^{\prime}(x) .\newlineAnswer: f′(x)= f^{\prime}(x)=
  1. Apply Chain Rule: To find the derivative of the function f(x)=2sec⁡(2x)f(x) = 2\sec(2x), we need to apply the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
  2. Identify Outer and Inner Functions: The outer function is sec⁡(u)\sec(u) where u=2xu = 2x, and the inner function is 2x2x. The derivative of sec⁡(u)\sec(u) with respect to uu is sec⁡(u)tan⁡(u)\sec(u)\tan(u), and the derivative of 2x2x with respect to xx is 22.
  3. Use Chain Rule Formula: Now we apply the chain rule: f′(x)=ddx[2sec⁡(2x)]=2⋅ddx[sec⁡(2x)]=2⋅sec⁡(2x)tan⁡(2x)⋅ddx[2x].f'(x) = \frac{d}{dx}[2\sec(2x)] = 2 \cdot \frac{d}{dx}[\sec(2x)] = 2 \cdot \sec(2x)\tan(2x) \cdot \frac{d}{dx}[2x].
  4. Substitute Inner Function Derivative: We already determined that ddx[2x]\frac{d}{dx}[2x] is 22, so we can substitute this into our expression: f′(x)=2⋅sec⁡(2x)tan⁡(2x)⋅2f'(x) = 2 \cdot \sec(2x)\tan(2x) \cdot 2.
  5. Simplify Expression: Simplify the expression by multiplying the constants: f′(x)=4sec⁡(2x)tan⁡(2x)f'(x) = 4\sec(2x)\tan(2x).

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