Question Statement

Find the derivative of the function:

$f(x)=(\sec 4x+\tan 2x{)}^5$

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Background and Explanation

This problem involves differentiating a composite function using the chain rule, combined with derivatives of trigonometric functions where the arguments are linear functions of $x$ (requiring additional chain rule applications).

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Solution

We need to find $f^{\prime }(x)$ where $f(x)=(\sec 4x+\tan 2x{)}^5$.

This is a composite function where the outer function is $u^5$ and the inner function is $u=\sec 4x+\tan 2x$.

Step 1: Apply the chain rule (power rule combined with chain rule):

$f^{\prime }(x)=5(\sec 4x+\tan 2x{)}^{5-1}\cdot \frac{d}{dx}(\sec 4x+\tan 2x)$

$f^{\prime }(x)=5(\sec 4x+\tan 2x{)}^4\cdot \frac{d}{dx}(\sec 4x+\tan 2x)$

Step 2: Differentiate the inner function term by term.

For $\frac{d}{dx}(\sec 4x)$:

• Using the chain rule: $\frac{d}{dx}[\sec u]=\sec u\tan u\cdot \frac{du}{dx}$ where $u=4x$
• $\frac{d}{dx}(\sec 4x)=\sec 4x\tan 4x\cdot \frac{d}{dx}(4x)=\sec 4x\tan 4x\cdot 4=4\sec 4x\tan 4x$

For $\frac{d}{dx}(\tan 2x)$:

• Using the chain rule: $\frac{d}{dx}[\tan u]={\sec }^{2}u\cdot \frac{du}{dx}$ where $u=2x$
• $\frac{d}{dx}(\tan 2x)={\sec }^{2}2x\cdot \frac{d}{dx}(2x)={\sec }^{2}2x\cdot 2=2{\sec }^{2}2x$

Step 3: Substitute these derivatives back into the expression:

$f^{\prime }(x)=5(\sec 4x+\tan 2x{)}^4\cdot \left(4\sec 4x\tan 4x+2{\sec }^{2}2x\right)$

Step 4: Factor out the common factor of 2 from the parentheses:

$f^{\prime }(x)=5(\sec 4x+\tan 2x{)}^4\cdot 2\left(2\sec 4x\tan 4x+{\sec }^{2}2x\right)$

Step 5: Simplify the numerical coefficients:

$f^{\prime }(x)=10(\sec 4x+\tan 2x{)}^4\cdot \left(2\sec 4x\tan 4x+{\sec }^{2}2x\right)$

Therefore, the derivative is:

$f^{\prime }(x)=10(\sec 4x+\tan 2x{)}^4\left(2\sec 4x\tan 4x+{\sec }^{2}2x\right)$

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Key Formulas or Methods Used

• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Power Rule with Chain Rule: $\frac{d}{dx}[u^n]=n{u}^{n-1}\cdot \frac{du}{dx}$
• Derivative of Secant: $\frac{d}{dx}[\sec u]=\sec u\tan u\cdot \frac{du}{dx}$
• Derivative of Tangent: $\frac{d}{dx}[\tan u]={\sec }^{2}u\cdot \frac{du}{dx}$
• Constant Multiple Rule: $\frac{d}{dx}[cu]=c\frac{du}{dx}$

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Summary of Steps

1. Identify the structure: Recognize $f(x)$ as a composite function (outer: power of 5, inner: trigonometric sum)
2. Apply chain rule: Differentiate the outer function $(\cdot {)}^5$ to get $5(\cdot {)}^4$, keeping the inner function unchanged
3. Differentiate inner terms:
   • For $\sec 4x$: apply chain rule to get $4\sec 4x\tan 4x$
   • For $\tan 2x$: apply chain rule to get $2{\sec }^{2}2x$
4. Combine results: Multiply the outer derivative by the sum of inner derivatives
5. Factor and simplify: Factor out 2 from the trigonometric terms and multiply by the existing 5 to get the final coefficient 10