Question Statement

Given the function:

$f(x)=\tan \frac{x}{2}$

Find the first derivative $f^{\prime }(x)$ and the second derivative $f^{\prime \prime }(x)$.

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Background and Explanation

This problem requires applying the chain rule for differentiation, which is essential when dealing with composite functions. You'll also need to recall the derivatives of basic trigonometric functions, specifically $\tan (u)$ and $\sec (u)$.

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Solution

Finding the First Derivative $f^{\prime }(x)$

We start with the given function:

$f(x)=\tan \frac{x}{2}$

Differentiate with respect to $x$ using the chain rule. Recall that $\frac{d}{du}(\tan u)={\sec }^{2}u$:

$\begin{array}{rl}{f}^{\prime }(x)& ={\sec }^2\frac{x}{2}\cdot \frac{d}{dx}\left(\frac{x}{2}\right)\\ & ={\sec }^2\frac{x}{2}\cdot \frac{1}{2}\\ & =\frac{1}{2}{\sec }^2\frac{x}{2}\end{array}$

Finding the Second Derivative $f^{\prime \prime }(x)$

Now we differentiate $f^{\prime }(x)=\frac{1}{2}{\sec }^2\frac{x}{2}$ with respect to $x$ again.

Apply the chain rule to ${\sec }^2\frac{x}{2}$:

• The outer function is $u^2$ (where $u=\sec \frac{x}{2}$), so its derivative is $2u=2\sec \frac{x}{2}$
• The middle function is $\sec v$ (where $v=\frac{x}{2}$), so its derivative is $\sec v\tan v=\sec \frac{x}{2}\tan \frac{x}{2}$
• The inner function is $\frac{x}{2}$, so its derivative is $\frac{1}{2}$

Putting it all together:

$\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{1}{2}\cdot \frac{d}{dx}\left({\sec }^2\frac{x}{2}\right)\\ & =\frac{1}{2}\cdot 2\sec \frac{x}{2}\cdot \frac{d}{dx}\left(\sec \frac{x}{2}\right)\\ & =\sec \frac{x}{2}\cdot \left(\sec \frac{x}{2}\tan \frac{x}{2}\right)\cdot \frac{d}{dx}\left(\frac{x}{2}\right)\\ & =\sec \frac{x}{2}\cdot \sec \frac{x}{2}\tan \frac{x}{2}\cdot \frac{1}{2}\\ & =\frac{1}{2}{\sec }^2\frac{x}{2}\tan \frac{x}{2}\end{array}$

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Key Formulas or Methods Used

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

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Summary of Steps

1. First Derivative: Apply the chain rule to $\tan (x/2)$:

   • Derivative of $\tan u$ is ${\sec }^{2}u$
   • Multiply by derivative of $x/2$ (which is $1/2$)
   • Result: $f^{\prime }(x)=\frac{1}{2}{\sec }^2(x/2)$

2. Second Derivative: Differentiate $\frac{1}{2}{\sec }^2(x/2)$ using chain rule:

   • Bring down the exponent: $2\sec (x/2)$
   • Multiply by derivative of $\sec (x/2)$: $\sec (x/2)\tan (x/2)\cdot (1/2)$
   • Include the constant $1/2$ from the first derivative
   • Simplify to get: $f^{\prime \prime }(x)=\frac{1}{2}{\sec }^2(x/2)\tan (x/2)$