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In Exercises \(1-18,\) find \(d y / d x\) $$ y=x^{2} \cos x-2 x \sin x-2 \cos x $$

Short Answer

Expert verified
\( \frac{dy}{dx} = -x^2 \sin x \)

Step by step solution

01

Identify the Function and Terms

Let's identify the terms in the given function. We have three distinct terms in the function: \( y = x^2 \cos x - 2x \sin x - 2\cos x \). Our goal is to compute the derivative \( \frac{dy}{dx} \).
02

Differentiate Each Term Individually

To find \( \frac{dy}{dx} \), differentiate each term of the function separately: 1. Differentiate \( x^2 \cos x \): - Use the product rule: \( u = x^2 \) and \( v = \cos x \). - Derivative of \( u \): \( u' = 2x \). - Derivative of \( v \): \( v' = -\sin x \). - Product rule: \((uv)' = u'v + uv' = 2x \cos x - x^2 \sin x \).2. Differentiate \(-2x \sin x\): - Use the product rule: \( u = -2x \) and \( v = \sin x \). - Derivative of \( u \): \( u' = -2 \). - Derivative of \( v \): \( v' = \cos x \). - Product rule: \((uv)' = u'v + uv' = -2\sin x - 2x\cos x \).3. Differentiate \(-2\cos x\): - Derivative is \( 2\sin x \) by the chain rule.
03

Combine the Derivatives

Now, combine all the derivatives obtained from Step 2:\[ \frac{dy}{dx} = (2x \cos x - x^2 \sin x) + (-2\sin x - 2x\cos x) + 2\sin x \].
04

Simplify the Expression

Simplify the derivative expression by combining like terms:- The \( 2x \cos x \) and \(-2x\cos x\) terms cancel out.- The \(-2\sin x\) and \(2\sin x\) terms also cancel each other.After simplification, we have:\[ \frac{dy}{dx} = -x^2 \sin x \].

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Key Concepts

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

Understanding the Product Rule
The Product Rule is a fundamental technique in calculus, especially useful for differentiating products of two functions. When you have a function in the form of the product of two expressions, such as \(f(x) = u(x) \cdot v(x)\), the Product Rule states that the derivative \(f'(x)\) is given by:
  • Derive the first function \(u(x)\): \(u'(x)\).
  • Multiply \(u'(x)\) with the second function \(v(x)\).
  • Derive the second function \(v(x)\): \(v'(x)\).
  • Multiply \(v'(x)\) with the first function \(u(x)\).
  • Add the two results: \(f'(x) = u'(x)\cdot v(x) + u(x)\cdot v'(x)\).

This rule becomes exceptionally handy when confronting expressions like \(x^2 \cos x\) or \(-2x \sin x\) in the original function. Always remember to apply this rule correctly to ensure that you calculate the correct expression for the derivatives. It allows you to break down complex derivatives into manageable parts, focusing on each function individually before combining them to form the complete derivative.
The Power of the Chain Rule
The Chain Rule is another crucial differentiation technique, particularly when dealing with composite functions or functions within functions. This rule is best described as applying when you take the derivative of a "nested" function, like \(h(x) = f(g(x))\). To differentiate such a function, you:
  • Take the derivative of the outer function, which we'll call \(f(x)\), at \(g(x)\), denoted as \(f'(g(x))\).
  • Take the derivative of the inner function \(g(x)\), which we'll call \(g'(x)\).
  • Multiply these derivatives: \(h'(x) = f'(g(x)) \cdot g'(x)\).

This approach is especially useful for trigonometric functions like \(-2\cos x\), where the derivative can be quickly found by recognizing that the derivative of \(\cos x\) is \(-\sin x\), and applying the Chain Rule if the argument of the function is more complex than a simple \(x\). This is a powerful way to handle more intricate expressions in calculus.
Working with Trigonometric Functions
Trigonometric functions are a staple in calculus, often integrated with the Product and Chain Rules to find derivatives. Common trig functions include \(\sin x\), \(\cos x\), and \(\tan x\), each having specific derivatives:
  • The derivative of \(\sin x\) is \(\cos x\).
  • The derivative of \(\cos x\) is \(-\sin x\).
  • The derivative of \(\tan x\) is \(\sec^2 x\).

These basic derivatives are cornerstones that form the basis for more complex derivative calculations involving trigonometric functions. When encountering a derivative problem, knowing these derivatives by heart will significantly aid in problem-solving. Whether you apply them directly or as part of the Product or Chain Rules, understanding trig derivatives simplifies many calculus exercises.

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Most popular questions from this chapter

Ships Two ships are steaming straight away from a point \(O\) along routes that make a \(120^{\circ}\) angle. Ship \(A\) moves at 14 knots (nautical miles per hour; a nautical mile is 2000 yd). Ship \(B\) moves at 21 knots. How fast are the ships moving apart when \(O A=5\) and \(O B=3\) nautical miles?

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