Differtentiat $$ x^{2}-\frac{1}{x^{2}} $$ $$ w \cdot x \cdot t $$

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Certainly! Let's differentiate each of the given functions with respect to $$ x $$. We'll assume that $$ w $$ and $$ t $$ are constants.

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### 1. Differentiate $$ f(x) = x^{2} - \frac{1}{x^{2}} $$

**Function:**
$$
f(x) = x^{2} - \frac{1}{x^{2}}
$$

**Differentiation:**
$$
\frac{d}{dx} \left( x^{2} \right) = 2x
$$
$$
\frac{d}{dx} \left( -\frac{1}{x^{2}} \right) = -\left( -2x^{-3} \right) = \frac{2}{x^{3}}
$$

**Result:**
$$
f'(x) = 2x + \frac{2}{x^{3}}
$$

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### 2. Differentiate $$ g(x) = w \cdot x \cdot t $$

**Function:**
$$
g(x) = w \cdot x \cdot t
$$
Since $$ w $$ and $$ t $$ are constants, $$ w \cdot t $$ can be treated as a single constant coefficient.

**Differentiation:**
$$
\frac{d}{dx} \left( w \cdot x \cdot t \right) = w \cdot t \cdot \frac{d}{dx} \left( x \right) = w \cdot t \cdot 1 = w t
$$

**Result:**
$$
g'(x) = w t
$$

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**Summary of Derivatives:**

1. $$\displaystyle \frac{d}{dx} \left( x^{2} - \frac{1}{x^{2}} \right) = 2x + \frac{2}{x^{3}} $$

2. $$\displaystyle \frac{d}{dx} \left( w \cdot x \cdot t \right) = w t $$

If you have any further questions or need additional clarification, feel free to ask!
The derivatives are: 1. $$ \frac{d}{dx} \left( x^{2} - \frac{1}{x^{2}} \right) = 2x + \frac{2}{x^{3}} $$ 2. $$ \frac{d}{dx} \left( w \cdot x \cdot t \right) = w t $$

Differtentiat \( x^{2}-\frac{1}{x^{2}} \) \( w \cdot x \cdot t \)

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