Question

Finding a Derivative In Exercises \(7-34,\) find the derivative of the function. $$ g(x)=\left(2+\left(x^{2}+1\right)^{4}\right)^{3} $$

Short answer

The derivative of \( g(x) = (2 + (x^{2}+1)^{4})^{3} \) is \( g'(x) = 24x(2 + (x^{2}+1)^{4})(x^{2}+1)^{3} \).

Step-by-step solution

Applying the chain rule

Begin the derivation by applying the chain rule. In simpler terms, differentiate the outer function keeping the inner function constant, then differentiate the inner function. The chain rule is written formally as: If \( g(x) = f(u), u=h(x) \), then \( g'(x)=f'(u).h'(x) \). For the given function, we have \( u = 2 + (x^{2}+1)^{4} \) and \( v = x^{2}+1 \). Apply the chain rule, which gives \( g'(x) = 3(2 + (x^{2}+1)^{4})^{2}.4(x^{2}+1)^{3}.2x \).

Simplify the derivative

Upon simplification of the above expression, we obtain \( g'(x) = 24x(2 + (x^{2}+1)^{4})(x^{2}+1)^{3} \).

Final derivative

Hence, the derivative of the function \( g(x) = (2 + (x^{2}+1)^{4})^{3} \) is \( g'(x) = 24x(2 + (x^{2}+1)^{4})(x^{2}+1)^{3} \).

Key concepts

Chain Rule

The chain rule is a fundamental technique in calculus used when differentiating compositions of functions. Imagine you have a function nested within another, much like a set of Russian dolls. The chain rule helps you "unwrap" these layers to find the derivative step-by-step.

When you have a function such as \( g(x) = f(u) \), where \( u = h(x) \), the chain rule instructs: differentiate \( f \) with respect to \( u \), then multiply by the derivative of \( u \) with respect to \( x \). Formally, this is expressed as:

• If \( g(x) = f(u) \) and \( u = h(x) \), then the derivative, \( g'(x) \), is given by \( f'(u) \cdot h'(x) \).

Consider how this rule manifests in the original problem where \( g(x) = (2 + (x^{2}+1)^{4})^{3} \). Here, we treat \( 2 + (x^{2}+1)^{4} \) as \( u \) and the entire expression as \( f(u) = u^3 \). Following the chain rule, the derivative involves computing the derivative of the outer function and then multiplying it by the derivative of the inner function.

Function Differentiation

Function differentiation is the process of finding the derivative of a function, which is the measure of how the function's output value changes as its input changes. Differentiation helps us understand the rate at which something is changing — often represented as the slope of a tangent to a curve.

When working with complex functions, differentiation can require several rules, one of them being the chain rule. In our problem, we first focus on differentiating the outermost layer, which is the power function. Given a term \( (2 + (x^{2}+1)^{4})^{3} \), differentiation with respect to the outer function leads to:

• Bring down the exponent: multiply the entire expression by 3.
• Subtract one from the exponent: raise the expression to the 2nd power.

This illustrates the straightforward application of power rule principles wrapped around the chain rule, reinforcing that each function must be meticulously peeled back like layers of an onion, revealing and differentiating each significant component of a composed function.

Simplification in Calculus

Simplification in calculus is the process of using algebraic techniques to make expressions easier to manage and understand. Simplification often involves combining like terms, factoring, and reducing complexities.

While deriving the original function's derivative, the expression can become cumbersome. That's where simplification steps in to transform a lengthy expression into something more concise and manageable. After applying the chain rule, the derivative expression \( 3(2 + (x^{2}+1)^{4})^{2} \cdot 4(x^{2}+1)^{3} \cdot 2x \) must be simplified.

• Combine constants into one multiplication term: \( 3 \times 4 \times 2 = 24 \).
• Redistribute these factors with the rest of the expression.

After simplification, the final derivative becomes much more straightforward: \( 24x(2 + (x^{2}+1)^{4})(x^{2}+1)^{3} \). This reduced form is easier to interpret and handle, providing clarity and aiding further calculations or assessments.