Question Statement

Find the fourth derivative $f^{(4)}(x)$ of the function:

$f(x)=\frac{1}{\sec (2x+1)}$

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Background and Explanation

This problem involves computing higher-order derivatives of trigonometric functions using the chain rule. First, simplify the function using reciprocal identities, then apply successive differentiation.

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Solution

First, simplify the function using the identity $\frac{1}{\sec \theta }=\cos \theta$:

$f(x)=\frac{1}{\sec (2x+1)}=\cos (2x+1)$

First Derivative:

Differentiate with respect to $x$ using the chain rule:

$\begin{array}{rl}{f}^{\prime }(x)& =-\sin (2x+1)\cdot \frac{d}{dx}(2x+1)\\ & =-\sin (2x+1)\cdot (2+0)\\ f^{\prime }(x)& =-2\sin (2x+1)\end{array}$

Second Derivative:

Differentiate again with respect to $x$:

$\begin{array}{rl}{f}^{\prime \prime }(x)& =-2\cos (2x+1)\cdot 2\\ f^{\prime \prime }(x)& =-4\cos (2x+1)\end{array}$

Third Derivative:

Differentiate again with respect to $x$:

$\begin{array}{rl}{f}^{\prime \prime \prime }(x)& =-4(-\sin (2x+1))\cdot 2\\ & =8\sin (2x+1)\end{array}$

Fourth Derivative:

Differentiate once more with respect to $x$:

$\begin{array}{rl}{f}^{(4)}(x)& =8\cdot \cos (2x+1)\cdot \frac{d}{dx}(2x+1)\\ & =8\cos (2x+1)\cdot 2\\ f^{(4)}(x)& =16\cos (2x+1)\end{array}$

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Key Formulas or Methods Used

• Reciprocal Identity: $\frac{1}{\sec \theta }=\cos \theta$
• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Trigonometric Derivatives:
  • $\frac{d}{dx}[\sin (u)]=\cos (u)\cdot u^{\prime }$
  • $\frac{d}{dx}[\cos (u)]=-\sin (u)\cdot u^{\prime }$
• Higher-Order Derivative Notation: $f^{(4)}(x)$ denotes the fourth derivative

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Summary of Steps

1. Simplify $f(x)=\frac{1}{\sec (2x+1)}$ to $\cos (2x+1)$
2. Apply chain rule to get $f^{\prime }(x)=-2\sin (2x+1)$
3. Differentiate to obtain $f^{\prime \prime }(x)=-4\cos (2x+1)$
4. Continue differentiation to find $f^{\prime \prime \prime }(x)=8\sin (2x+1)$
5. Final differentiation yields $f^{(4)}(x)=16\cos (2x+1)$