Problem 94 - Calculus and its applications

Question

Differentiate. $$ F(x)=\left[6 x(3-x)^{5}+2\right]^{4} $$

Step-by-Step Solution

Answer

Apply the chain rule: \(F'(x) = 4[6x(3-x)^5 + 2]^3 \cdot [-24x + 18](3-x)^4\).

Step 1: Identify the Rule to Use

The function \(F(x) = [6x(3-x)^{5} + 2]^{4}\) is a composition of multiple functions, specifically, a chain of a polynomial inside a power function. To differentiate it, we need to apply the chain rule, where if \(y = [u(x)]^n\), then \(\frac{dy}{dx} = n[u(x)]^{n-1} \cdot \frac{du}{dx}\).

Step 2: Differentiate the Outer Function

Let \(u(x) = 6x(3-x)^{5} + 2\). According to the chain rule, differentiate the outer function \((u(x))^{4}\). This gives us: \(4[u(x)]^{3}\) since the derivative of \(v^n\) with respect to \(v\) is \(n \cdot v^{n-1}\).

Step 3: Compute the Derivative of the Inner Function

Differentiate the inner function \(u(x) = 6x(3-x)^{5} + 2\). To differentiate \(6x(3-x)^5\), apply the product rule, where if \(y = uv\), then \(\frac{dy}{dx} = u'v + uv'\). First, express it as \(u = 6x\) and \(v = (3-x)^5\).

Step 4: Differentiate using the Product Rule

For \(u = 6x\), the derivative is \(6\). For \(v = (3-x)^5\), use the power and chain rule: \(-5(3-x)^4\) (since the derivative of \(3-x\) is \(-1\)). Now apply the product rule: \(6(3-x)^5 + 6x[-5(3-x)^4]\).

Step 5: Simplify the Inner Derivative

Combine the terms from the product rule: \(6(3-x)^5 - 30x(3-x)^4\). So, the derivative of the inside function \(u(x)\) is: \((3-x)^4[6(3-x) - 30x]\). Simplify further if needed.

Step 6: Apply the Chain Rule

Now that we have \(u'(x)\), plug it into the chain rule equation: \(F'(x) = 4[6x(3-x)^5 + 2]^{3} \cdot [(3-x)^4(6(3-x) - 30x)]\).

Step 7: Final Simplification

Substitute and simplify: \(F'(x) = 4[6x(3-x)^5 + 2]^3 \cdot [-24x + 18](3-x)^4\).

Key Concepts

Chain Rule

The chain rule is an essential technique in calculus used to differentiate composite functions. When a function is embedded within another, like in nested layers, the chain rule helps peel away each layer one at a time. It can almost be thought of as an 'inner' and 'outer' function process.

For example, if we have a function like \(F(x) = [u(x)]^n\), the chain rule tells us to first differentiate the outer layer, thinking of \(u(x)\) as a single variable. So, we differentiate \(v^n\) with respect to \(v\), which gives \(n \/ v^{n-1}\).

Next, we don't forget about the inner function \(u(x)\). We've differentiated the outer layer, and it's now time to multiply the result by the derivative of the inner function, \( rac{du}{dx}\). By connecting these pieces, we apply the chain rule to successfully differentiate composite functions.

Product Rule

The product rule is a tool used when differentiating a product of two functions. When you see a function that is a multiplication of two separate parts, this is where the product rule shines.

The general formula is \( rac{d}{dx}(u \, v) = u'v + uv' \), which means that you differentiate one component at a time while always keeping the other component in its original form. Then, add the two results.

For instance, if you have \(y = 6x(3-x)^5\), break it into two parts, \(u = 6x\) and \(v = (3-x)^5\). Now, apply the rule: differentiate \(u\), giving \(6\), and differentiate \(v\) using the power and chain rule, yielding \(-5(3-x)^4\). Combine these using the product rule: \(6(3-x)^5 + 6x[-5(3-x)^4]\), allowing for further simplification.

Composite Functions

Composite functions involve one function nested, or composed, inside another. This means the output of one function becomes the input of another. When dealing with differentiation, they can appear complicated, but following step-by-step rules like the chain rule makes them approachable.

An example of a composite function is \(F(x) = [6x(3-x)^5 + 2]^4\). You can see the layering: it's a polynomial \(6x(3-x)^5 + 2\) within a fourth power.

Understanding composite functions requires seeing them in layers. Focus on identifying their inner and outer functions and handle differentiation in a methodical manner using the chain rule. This approach allows breaking down the problem into manageable parts, making differentiation of even complex compositions straightforward.