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Chapter 3: Problem 60

Suppose that \(y\) is a differentiable function of \(x\). Express the derivative of the given function with respect to \(x\) in terms of \(x, y\), and \(d y / d x\). \(\cos x^{5} y^{3}\)

Short Answer

Expert verified
The derivative is \(-\sin(x^5 y^3) \cdot (5x^4 y^3 + 3x^5 y^2 \frac{dy}{dx})\)."

Step by step solution

01

Identify the function components

The given function is \( \cos(x^5 y^3) \). Inside the cosine function, we have the inner function \( u = x^5 y^3 \). This is a composition of functions, so we'll use the chain rule for differentiation.
02

Differentiate the outer function

Using the chain rule, the derivative of \( \cos u \) is \( -\sin u \), where \( u = x^5 y^3 \). Thus, \( \frac{d}{dx}(\cos u) = -\sin(u) \cdot \frac{du}{dx} \). Now, we need to find \( \frac{du}{dx} \).
03

Differentiate the inner function using product rule

The inner function is \( u = x^5 y^3 \), which is a product of two functions \( x^5 \) and \( y^3 \). Differentiating using the product rule: \( \frac{du}{dx} = \frac{d}{dx}(x^5) \cdot y^3 + x^5 \cdot \frac{d}{dx}(y^3) \).
04

Apply differentiation

Differentiating \( x^5 \) with respect to \( x \) gives \( 5x^4 \). Differentiating \( y^3 \) with respect to \( x \) requires the chain rule: \( 3y^2 \cdot \frac{dy}{dx} \). Thus, \( \frac{du}{dx} = 5x^4 y^3 + x^5 \cdot 3y^2 \frac{dy}{dx} \).
05

Substitute back into the expression

Substituting \( \frac{du}{dx} \) into the chain rule expression: \(-\sin(x^5 y^3) \cdot (5x^4 y^3 + 3x^5 y^2 \frac{dy}{dx}) \). This is the derivative of \( \cos(x^5 y^3) \) with respect to \( x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is a powerful technique in calculus used for differentiating composed functions. It's like peeling layers of an onion; you work with each layer individually. When you have a function inside another function, such as \(\cos(x^5 y^3)\), the chain rule helps to find the derivative.
  • First, identify the outer and inner functions. Here, \(\cos(\cdot)\) is the outer function and \(x^5 y^3\) is the inner one.
  • The derivative according to the chain rule is: \(\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)\).
  • In our example, this becomes \(-\sin(x^5 y^3) \cdot \frac{d}{dx}(x^5 y^3)\).
Remember, the chain rule is all about multiplying the derivative of the outer function by the derivative of the inner function, evaluated at the given point.
Differentiation
Differentiation is a fundamental operation in calculus that finds the rate of change of a function. It's like asking how quickly one quantity changes with respect to another. For the given problem, differentiation helps determine how the function \(\cos(x^5 y^3)\) changes as \(x\) changes.
  • Use basic rules like the power rule which says \(\frac{d}{dx}(x^n) = nx^{n-1}\).
  • The function \(x^5\) differentiates to \(5x^4\).
  • For a function like \(y^3\), which seems constant in relation to \(x\), use implicit differentiation: \(3y^2 \cdot \frac{dy}{dx}\).
Differentiation allows us to peel apart the function and look at each component's rate of change separately.
Product Rule
The product rule applies when you're differentiating a product of two functions, a situation you'll frequently encounter. It's like having two gears interlocking; if one moves, it affects the movement of the other. Let's break down its application in our function \(x^5 y^3\).
  • The product rule formula is: \((uv)' = u'v + uv'\), where \(u\) and \(v\) are functions of \(x\).
  • In this context, \(u = x^5\) and \(v = y^3\).
  • Differentiate each part separately: \(\frac{d}{dx}(x^5) = 5x^4\) and \(\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}\).
  • The result is \(5x^4 y^3 + x^5 \cdot 3y^2 \cdot \frac{dy}{dx}\).
The product rule lets us tackle the complexity of differentiating products by dealing with each term as if isolated, while piecing them back together at the end.

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Most popular questions from this chapter

If \(A, x\), and \(h\) denote the area, length of side, and altitude of an equilateral triangle, respectively, then they are related by the formulas $$ A=\frac{\sqrt{3}}{4} x^{2} \quad \text { and } \quad x=\frac{2 \sqrt{3}}{3} h $$ Using the Chain Rule, find the rate of change of the area with respect to the altitude, and determine the rate of change when \(h=\sqrt{3}\).

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Use the Newton-Raphson method to find an approximate solution of the given equation in the given interval. Use the method until successive approximations obtained by calculator are identical. $$ e^{-x}=x ;[0,1] $$
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