Question

Compute the derivative of the given function. $$p(t)=\cos ^{3}\left(t^{2}+3 t+1\right)$$

Short answer

The derivative is \( -3(\cos(t^2 + 3t + 1))^2 \sin(t^2 + 3t + 1)(2t + 3) \).

Step-by-step solution

Recognize the Composition of Functions

The function given is a composition of two functions: an outer function \( f(u) = u^3 \) and an inner function \( u = \cos(v) \), where \( v = t^2 + 3t + 1 \). To find the derivative, we will use the chain rule multiple times.

Apply the Chain Rule to the Outer Function

First, differentiate the outer function with respect to \( u \). If \( p(u) = u^3 \), then \( p'(u) = 3u^2 \). So, if \( u = \cos(v) \), this becomes \( 3(\cos(v))^2 \).

Differentiate the Inner Function

Differentiate \( u = \cos(v) \) with respect to \( v \). The derivative of \( \cos(v) \) is \(-\sin(v) \).

Differentiate the Innermost Function

Next, differentiate \( v = t^2 + 3t + 1 \) with respect to \( t \). The derivative is \( 2t + 3 \).

Combine Using Chain Rule

Combine these derivatives using the chain rule for function composition:\[ \frac{dp}{dt} = \frac{dp}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dt} = 3(\cos(v))^2 \cdot (-\sin(v)) \cdot (2t + 3) \]

Substitute Back for v

Substitute back \( v = t^2 + 3t + 1 \) into the equation:\[ \frac{dp}{dt} = 3(\cos(t^2 + 3t + 1))^2 \cdot (-\sin(t^2 + 3t + 1)) \cdot (2t + 3) \]

Simplify the Expression

The expression can be simplified by combining terms:\[ \frac{dp}{dt} = -3(\cos(t^2 + 3t + 1))^2 \cdot \sin(t^2 + 3t + 1) \cdot (2t + 3) \]

Key concepts

Chain Rule

The chain rule is an essential concept when differentiating compositions of functions. It tells us how to take the derivative of a composite function, which is when one function is nested inside another function. The typical formulation is:

• If you have a function composed as \( f(g(x)) \), then its derivative is \( f'(g(x)) \cdot g'(x) \).
• This process involves differentiating the outer function while leaving the inner function unchanged and then multiplying by the derivative of the inner function.

In our exercise, applying the chain rule multiple times allowed us to break down the function \( p(t) = \cos^3(t^2 + 3t + 1) \) by differentiating each layer step-by-step. This systematic approach makes complex derivatives much more manageable.

Composition of Functions

The composition of functions involves two (or more) functions being united to form a single new function. It's like stacking actions one inside another. Consider this example:

• First, take a simple function \( u = \cos(v) \) where \( v = t^2 + 3t + 1 \).
• The outer function is taking a cubed power, effectively making it \( \cos^3(v) \).
• Each function (inner and outer) maintains its own derivative rules, which are critical in applying when using the chain rule.

Understanding compositions makes it easier to dissect complex functions like the one given and apply the appropriate differentiation techniques needed to find derivatives. It's the layering of functions that creates both a structure and a path to follow for differentiation.

Trigonometric Functions

Trigonometric functions like cosine and sine are fundamental in calculus, particularly when combined with other operations. Here's what to keep in mind:

• The cosine function, represented as \( \cos(x) \), has a derivative of \( -\sin(x) \). This negative sign is crucial to remember.
• These functions frequently appear in compositions where they can be raised to powers or taken as inputs for other functions.
• In the problem, we see \( \cos(v) \) being cubed, indicating that both the power rule and trigonometric concepts are in play.

Recognizing these patterns allows you to efficiently apply derivative rules and transform a trigonometric function into something approachable and computable using known derivative operations. They are integral in navigating the derivative landscape within calculus.