Problem 13 - The Calculus with Analytic Geometry

Question

Find the derivative of the given function. $$ h(x)=\frac{\sqrt{x-1}}{\sqrt[3]{x+1}} $$

Step-by-Step Solution

Answer

The derivative of h(x) is \[ h'(x) = \frac{x+4}{6(x-1)^{1/2}(x+1)^{5/3}}. \]

Step 1: Rewrite the Function

Rewrite the function oindent \[ h(x) = \frac{\sqrt{x-1}}{\sqrt[3]{x+1}} \] in terms of exponents for easier differentiation. Recall that oindent \[ \sqrt{x-1} = (x-1)^{1/2} \] and oindent \[ \sqrt[3]{x+1} = (x+1)^{1/3} \], thus oindent \[ h(x) = \frac{(x-1)^{1/2}}{(x+1)^{1/3}}. \]

Step 2: Apply Quotient Rule

Use the quotient rule for differentiation which states \[ \left(\frac{u}{v}\right)' = \frac{vu' - uv'}{v^2} \]. Let oindent \[ u = (x-1)^{1/2} \] and oindent \[ v = (x+1)^{1/3} \].

Step 3: Find the Derivatives of u and v

Find oindent \[ u' = \frac{d}{dx}(x-1)^{1/2} \]. Using the chain rule, \[ u' = \frac{1}{2}(x-1)^{-1/2} \] and \[ v' = \frac{d}{dx}(x+1)^{1/3} \] using the chain rule, \[ v' = \frac{1}{3}(x+1)^{-2/3} \].

Step 4: Apply Quotient Rule Formula

Substitute oindent \[ u, u', v, \] and \[ v' \] into the quotient rule formula: \[ h'(x) = \frac{(x+1)^{1/3} \cdot \frac{1}{2}(x-1)^{-1/2} - (x-1)^{1/2} \cdot \frac{1}{3}(x+1)^{-2/3}}{[(x+1)^{1/3}]^2}. \]

Step 5: Simplify the Expression

Simplify the expression: \[ h'(x) = \frac{(x+1)^{1/3}}{2(x-1)^{1/2}(x+1)^{2/3}} - \frac{(x-1)^{1/2}}{3(x+1)^{5/3}}. \] Combine the fractions over a common denominator: \[ h'(x) = \frac{3(x+1)(x+1)^{-1} - 2(x-1)}{6(x-1)^{1/2}(x+1)^{5/3}}. \] Finally, further simplification gives \[ h'(x) = \frac{x+4}{6(x-1)^{1/2}(x+1)^{5/3}}. \]

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental concept in calculus used to find the derivative of a quotient of two functions. It states that if you have a function ] '' is equal to ''...'' minus ''...'' times ''...'' squared. This seems complicated, but allows us to differentiate quotients with ease.

Chain Rule

The Chain Rule is another essential rule in calculus used for differentiating composite functions. In simple terms, it states that to find the derivative of a composite function, you take the derivative of the outer function and multiply it by the derivative of the inner function. For example, for a function \(f(g(x))\), the derivative is \(f \text { ' } (g(x))g \text { ' } (x).\)
In our exercise, we used the chain rule twice, once for \(u = (x-1)^{1/2}\) and once for \(v = (x+1)^{1/3}\).$$Recap; the chain rule is used for finding the derivative of more complicated functions quickly and accurately.

Simplification in Calculus

Simplification is a crucial part of solving any calculus problem. It involves reducing expressions to their simplest forms to make solving them easier and more manageable. In the given exercise, after applying the quotient and chain rules, we simplified the resultant expression: Combining fractions involves finding a common denominator and then simplifying the terms. Finally, the expression becomes clear and easier to understand. This shows why simplification is important in calculus.