Problem 12 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\sqrt{x^{3}-2 x+5} $$

Step-by-Step Solution

Answer

The derivative of the function is \(f'(x) = \frac{3x^2 - 2}{2\sqrt{x^3 - 2x + 5}}\).

Step 1: Identify the Function Type

The given function is a composite function, where the outer function is a square root and the inner function is a polynomial. We will need to use the Chain Rule to differentiate it.

Step 2: Apply the Chain Rule

The Chain Rule states that if you have a composite function, say \(f(g(x))\), the derivative is \(f'(g(x)) \cdot g'(x)\). For our function, set \(u = x^3 - 2x + 5\) so that \(f(x) = \sqrt{u}\).

Step 3: Differentiate the Outer Function

Differentiate the outer function \(\sqrt{u}\) with respect to \(u\). The derivative of \(\sqrt{u}\) is \(\frac{1}{2}u^{-\frac{1}{2}}\) or \(\frac{1}{2\sqrt{u}}\).

Step 4: Differentiate the Inner Function

Now differentiate the inner function \(u = x^3 - 2x + 5\) with respect to \(x\). The derivative is \(u'(x) = 3x^2 - 2\).

Step 5: Apply the Chain Rule

Using the chain rule, multiply the derivative of the outer function by the derivative of the inner function: \(f'(x) = \frac{1}{2\sqrt{x^3 - 2x + 5}} \cdot (3x^2 - 2)\).

Step 6: Simplify the Expression

The derivative \(f'(x)\) simplifies to \(\frac{3x^2 - 2}{2\sqrt{x^3 - 2x + 5}}\).

Key Concepts

Understanding Composite Functions

A composite function is essentially a combination of two or more functions, where one function is applied to the result of another. In the given exercise, the function \( f(x) = \sqrt{x^3 - 2x + 5} \) is a clear example of a composite function.
The outer function here is the square root, \( f(u) = \sqrt{u} \), and the inner function is the polynomial, \( u(x) = x^3 - 2x + 5 \).

• The square root function processes the result of the polynomial.
• This nesting of functions means special rules must be applied when differentiating, such as the Chain Rule.

The key is to first identify the different layers of your function, being mindful of which are outer and which are inner. Once these are clear, we apply differentiation rules accordingly.

The Derivative and Chain Rule

The derivative tells us how a function changes as its input changes. In calculus, the Chain Rule is a principle that simplifies the calculation of the derivative of a composite function.
For example, with a composite like \(f(g(x))\), the Chain Rule states:

• Differentiate the outer function, evaluated at the inner function: \(f'(g(x))\).
• Multiply this by the derivative of the inner function, \(g'(x)\).

When we apply the Chain Rule for our original exercise, we are first finding the derivative of \( \sqrt{u} \) with respect to \( u\), resulting in \( \frac{1}{2\sqrt{u}} \).
Then we find the derivative of \( x^3 - 2x + 5 \) with respect to \( x \), which is \( 3x^2 - 2 \).
Finally, we multiply these results together to get the overall derivative. This method neatly handles the complexity of differentiating each part of the composite function.

Working with Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. In the exercise, the polynomial function is \( x^3 - 2x + 5 \).
Polynomial functions are straightforward when it comes to differentiation:

• The derivative of \( x^n \) is \( nx^{n-1} \).
• For example, differentiating \( x^3 \) gives \( 3x^2 \).
• Constants, like \( +5 \), have a derivative of zero.

Combining these principles, the inner function \( x^3 - 2x + 5 \) is differentiated to \( 3x^2 - 2 \).
This simple process makes polynomial functions a staple of calculus, offering clear, manageable rules for computation.