Question

86-89. Second derivatives Find \(\frac{d^{2} y}{d x^{2}}\) for the following functions. $$y=\sqrt{x^{2}+2}$$

Short answer

Answer: The second derivative of the function is \(\frac{d^2y}{dx^2} = \frac{2}{(x^2+2)\sqrt{x^2+2}}\).

Step-by-step solution

Find the first derivative of \(y\) with respect to \(x\)

Recall the chain rule for differentiation: if \(u(x)\) is a function of \(x\), then the derivative of \(u(x)\) with respect to \(x\) is given by $$\frac{d(u(x))}{dx} = \frac{du(x)}{dx} \cdot \frac{d(x^2+2)}{dx}.$$ Here, the function \(y=\sqrt{x^2+2}\) can be written as \(y=u(x^2+2)\) with \(u(x)=\sqrt{x}\). Now, let's find the first derivative \(\frac{dy}{dx}\): $$\frac{dy}{dx} = \frac{d}{dx} \sqrt{x^2+2} = \frac{d(\sqrt{t})}{dt} \cdot \frac{d(x^2+2)}{dx},$$ where we let \(t = x^2+2\). As per the chain rule, we have $$\frac{dy}{dx} = \frac{1}{2\sqrt{t}} \cdot (2x) = \frac{x}{\sqrt{x^2+2}}.$$

Differentiate \(\frac{dy}{dx}\) once more

Now, we need to differentiate the first derivative, \(\frac{dy}{dx}\), to obtain the second derivative, \(\frac{d^2y}{dx^2}\). For that, we will use the quotient rule: The quotient rule formula states that given a function of the form \(\frac{u(x)}{v(x)}\), the derivative of the function with respect to \(x\) is given by $$\frac{d}{dx} \frac{u(x)}{v(x)} = \frac{u'(x)v(x) - u(x)v'(x)}{v^2(x)}.$$ Applying the quotient rule to \(y'(x) = \frac{x}{\sqrt{x^2+2}}\), we obtain the second derivative: $$\frac{d^2y}{dx^2} = \frac{d}{dx} \frac{x}{\sqrt{x^2+2}} = \frac{(\sqrt{x^2+2} \cdot 1) - (x \cdot \frac{x}{\sqrt{x^2+2}})}{(x^2+2)}.$$ Simplify the expression: $$\frac{d^2y}{dx^2} = \frac{\sqrt{x^2+2} - \frac{x^2}{\sqrt{x^2+2}}}{(x^2+2)}.$$

Express the second derivative in a simplified form

Further simplification can be done by multiplying the numerator and denominator by the conjugate of the numerator, which is \(\sqrt{x^2+2} + \frac{x^2}{\sqrt{x^2+2}}\). Doing so, we get: $$\frac{d^2y}{dx^2} = \frac{(\sqrt{x^2+2} - \frac{x^2}{\sqrt{x^2+2}})(\sqrt{x^2+2} + \frac{x^2}{\sqrt{x^2+2}})}{(x^2+2)(\sqrt{x^2+2} + \frac{x^2}{\sqrt{x^2+2}})}.$$ Simplify the expression: $$\frac{d^2y}{dx^2} = \frac{(x^2+2) - x^2}{(x^2+2)(\sqrt{x^2+2})} = \frac{2}{(x^2+2)\sqrt{x^2+2}}.$$ Finally, the second derivative of \(y=\sqrt{x^2+2}\) is: $$\frac{d^2y}{dx^2} = \frac{2}{(x^2+2)\sqrt{x^2+2}}.$$

Key concepts

Chain Rule

The chain rule is an essential tool in calculus for differentiating compositions of functions. It provides a method for finding the derivative of a composite function. Imagine you have a function nested inside another function, just like a Russian doll—a doll within a doll.
When working with such functions, simply differentiating them isn't straightforward. This is where the chain rule shines!

• The chain rule states: if you have a function \( y = f(g(x)) \), the derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
• Essentially, you differentiate the outer function and multiply it by the derivative of the inner function.

In our exercise, the function \( y = \sqrt{x^2+2} \) can be rewritten using another variable \( t \), where \( t = x^2 + 2 \) and \( y = \sqrt{t} \).
By substituting this, differentiating becomes simpler using the chain rule:

• The outer function is \( \sqrt{t} \), differentiating gives \( \frac{1}{2\sqrt{t}} \).
• The inner function is \( x^2+2 \), with a derivative of \( 2x \).

Combining these results, according to the chain rule, the first derivative \( \frac{dy}{dx} \) becomes \( \frac{x}{\sqrt{x^2+2}} \).
This process breaks down the differentiation into manageable parts, making it straightforward, yet comprehensive.

First Derivative

The first derivative of a function reveals how the function changes at any given point—essentially, its slope at various points. It provides insights about the rate at which the function increases or decreases, a vital tool in understanding the function's behavior.

• For our function \( y = \sqrt{x^2+2} \), finding the first derivative is pivotal to understanding the initial rate of change.
• Using the setup from the previous section, we apply the chain rule to differentiate \( y \).

The resulting derivative \( \frac{dy}{dx} = \frac{x}{\sqrt{x^2+2}} \) gives detailed information about how changes in \( x \) affect the function.
For instance, as \( x \) changes, the derivative shows whether the graph of \( y \) is increasing or decreasing.
This foundational step is crucial when seeking the second derivative, as it lays the groundwork for further exploration.

• Knowing \( \frac{dy}{dx} \) helps in determining maximum or minimum values and understanding the curvature of \( y \).

Overall, the first derivative sets the stage for uncovering more complex attributes of the function.

Quotient Rule

In calculus, situations often arise where you need to differentiate a fraction of two functions. The quotient rule helps handle these scenarios beautifully. Specifically designed for fractions, it ensures that even complex derivatives are within reach.
The quotient rule states: for a function \( \frac{u(x)}{v(x)} \), the derivative is \[ \frac{d}{dx} \frac{u(x)}{v(x)} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \].

• This formula helps differentiate expressions like \( \frac{x}{\sqrt{x^2+2}} \) from our exercise.

We apply the quotient rule to find the second derivative, \( \frac{d^2y}{dx^2} \), by first considering \( u(x) = x \) and \( v(x) = \sqrt{x^2+2} \).

• Using the quotient rule, we differentiate \( y' \) to obtain \( \frac{d^2y}{dx^2} \).
• The process involves plugging into the formula, simplifying step-by-step.

After applying the quotient rule and simplifying, we reach the expression for the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{2}{(x^2+2)\sqrt{x^2+2}} \].
This complete journey—from using the chain rule to find \( y' \), to applying the quotient rule for \( y'' \)—demonstrates the power and necessity of these rules in unraveling the behavior of functions.