Problem 9 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 10, find the first and second derivative of the function defined by the given equation. $$ G(x)=\frac{1}{\sqrt{3+2 x^{2}}} $$

Step-by-Step Solution

Answer

The first derivative is \( G'(x) = -2x (3 + 2x^2)^{-\frac{3}{2}} \). The second derivative is \( G''(x) = -2 (3 + 2x^2)^{-\frac{3}{2}} + 12x^2 (3 + 2x^2)^{-\frac{5}{2}} \).

Step 1 - Rewrite the function in a more convenient form

Rewrite the function to a form that is easier to differentiate. Given the function \( G(x) = \frac{1}{\sqrt{3 + 2x^2}} \), we can rewrite this as \( G(x) = (3 + 2x^2)^{-\frac{1}{2}} \).

Step 2 - Find the first derivative using the chain rule

Apply the chain rule. Start by differentiating the outer function. For \( G(x) = (3 + 2x^2)^{-\frac{1}{2}} \), let \( u = 3 + 2x^2 \). Then, \( G(x) = u^{-\frac{1}{2}} \). The derivative of \( u^{-\frac{1}{2}} \) with respect to \( u \) is \(-\frac{1}{2} u^{-\frac{3}{2}} \). The derivative of \( u \) with respect to \( x \) is \( 4x \). Therefore, the first derivative is: \[ G'(x) = -\frac{1}{2} (3 + 2x^2)^{-\frac{3}{2}} \times 4x = -2x (3 + 2x^2)^{-\frac{3}{2}} \].

Step 3 - Find the second derivative

Now, differentiate the first derivative \( G'(x) \) again using the chain rule and product rule. Starting from \( G'(x) = -2x (3 + 2x^2)^{-\frac{3}{2}} \), let \( v = (3 + 2x^2)^{-\frac{3}{2}} \) and use the product rule: \((uv)' = u'v + uv'\). Here, \( u = -2x \) and \( u' = -2 \), \( v = (3 + 2x^2)^{-\frac{3}{2}} \), and \( v' = -\frac{3}{2} (3 + 2x^2)^{-\frac{5}{2}} \times 4x = -6x (3 + 2x^2)^{-\frac{5}{2}} \). Hence, \[ G''(x) = -2 (3 + 2x^2)^{-\frac{3}{2}} + (-2x) \times (-6x (3 + 2x^2)^{-\frac{5}{2}}) = -2 (3 + 2x^2)^{-\frac{3}{2}} + 12x^2 (3 + 2x^2)^{-\frac{5}{2}}. \]

Key Concepts

chain rule

The chain rule is a vital differentiation technique in calculus used when dealing with composite functions. A composite function is a function made up of other functions. For example, if we have a function defined as \( G(x) = (3 + 2x^2)^{-\frac{1}{2}} \), this can be seen as the composition of two functions: an outer function \( f(u) = u^{-\frac{1}{2}} \) and an inner function \( u = 3 + 2x^2 \).

The chain rule tells us to differentiate the outer function first, then multiply it by the derivative of the inner function. So, the outer function derivative \( \frac{d}{du}(u^{-\frac{1}{2}}) \) results in \( -\frac{1}{2} u^{-\frac{3}{2}} \). Next, we differentiate the inner function \( u = 3 + 2x^2 \), giving us \( \frac{d}{dx}(3 + 2x^2) = 4x \). Finally, applying the chain rule, we multiply these derivatives together:
\text{First derivative:} \[ \frac{d}{dx}(3 + 2x^2)^{-\frac{1}{2}} = -\frac{1}{2}(3 + 2x^2)^{-\frac{3}{2}} \times 4x = -2x(3 + 2x^2)^{-\frac{3}{2}} \] Breaking tasks this way simplifies differentiation.

product rule

When a function is a product of two or more functions, the product rule comes into play. The product rule states that for any two functions \( u(x) \) and \( v(x) \), the derivative of their product is given by \( (uv)' = u'v + uv' \). Let's see this in action, using the first derivative we found earlier:
- Let \( u = -2x \) and \( v = (3 + 2x^2)^{-\frac{3}{2}} \).
\text{First, find the individual derivatives:}
- \( u' = -2 \) and \( v' = -\frac{3}{2}(3 + 2x^2)^{-\frac{5}{2}} \times 4x = -6x (3 + 2x^2)^{-\frac{5}{2}} \).
Now, apply the product rule:
\text{Second derivative:} \[ G''(x) = -2(3 + 2x^2)^{-\frac{3}{2}} + (-2x)(-6x(3 + 2x^2)^{-\frac{5}{2}}) = -2(3 + 2x^2)^{-\frac{3}{2}} + 12x^2(3 + 2x^2)^{-\frac{5}{2}} \]
It shows how powerful the product rule is, especially when combined with the chain rule for more complex functions.

differentiation

Differentiation is the process of finding the derivative of a function, representing the rate of change of the function with respect to its variable. It is a fundamental concept in calculus. Here's a quick recap:

• To differentiate a function means finding its derivative.
• The derivative, represented as \( f'(x) \), measures how the function \( f(x) \) changes as \( x \) changes.
• The differentiation rules, such as the chain rule and product rule, simplify the process.

Consider the function given: \[ G(x) = \frac{1}{\text{sqrt}(3 + 2x^2)} \], we re-wrote it as \[ G(x) = (3 + 2x^2)^{-\frac{1}{2}} \]. Using the chain rule to differentiate gave us the first derivative: \[ G'(x) = -2x(3 + 2x^2)^{-\frac{3}{2}} \].

Then, we applied the product rule to find the second derivative: \[ G''(x) = -2(3 + 2x^2)^{-\frac{3}{2}} + 12x^2(3 + 2x^2)^{-\frac{5}{2}} \].
Differentiation allows us to find these rates of change effectively. Mastering differentiation techniques like the chain rule and product rule is key to solving complex problems in calculus.