Problem 1
Question
Find \(d y / d x\) $$ y=-10 x+3 \cos x $$
Step-by-Step Solution
Verified Answer
The derivative is \\(\frac{d y}{d x} = -10 - 3 \sin x\\).
1Step 1: Differentiating the Linear Term
Identify and differentiate the linear term in the function. The term is \(-10x\), where the derivative is calculated using the power rule, yielding \-10\.
2Step 2: Differentiating the Trigonometric Term
Identify and differentiate the trigonometric term involving cosine. The term is \(+3 \cos x\). Use the derivative of \(\cos x\), which is \(-\sin x\), and apply the chain rule resulting in \(-3 \sin x\).
3Step 3: Combining the Derivatives
Combine the derivatives of each term to get the final expression. The derivative of the entire function is the sum of the individual derivatives: \(-10 - 3 \sin x\).
Key Concepts
Power Rule and Its Use in DifferentiationTrigonometric Differentiation and Its ImportanceChain Rule in Differentiation
Power Rule and Its Use in Differentiation
The power rule is a key concept in differentiation and is one of the simplest rules to apply. It states that for any function of the form \(f(x) = ax^n\), where \(a\) is a constant and \(n\) is a real number, the derivative \(f'(x)\) is given by \(a \cdot n \cdot x^{n-1}\).
For example, when differentiating \(-10x\), you treat \(-10\) as \(a\) and the power of \(x\) as 1. Applying the power rule, the derivative is \(-10 \times 1 \times x^{1-1} = -10\).
This concept is particularly useful for quickly finding derivatives of simple polynomials. It simplifies the process by reducing the required calculation to a straightforward multiplication and subtraction.
For example, when differentiating \(-10x\), you treat \(-10\) as \(a\) and the power of \(x\) as 1. Applying the power rule, the derivative is \(-10 \times 1 \times x^{1-1} = -10\).
This concept is particularly useful for quickly finding derivatives of simple polynomials. It simplifies the process by reducing the required calculation to a straightforward multiplication and subtraction.
Trigonometric Differentiation and Its Importance
Trigonometric differentiation involves finding the derivative of functions that contain trigonometric terms like \(\sin(x)\), \(\cos(x)\), etc. These derivatives rely on known results:
This technique helps in tackling complex problems that incorporate trigonometric terms, enabling a clearer and more organized approach to solving derivatives.
- The derivative of \(\sin(x)\) is \(\cos(x)\).
- The derivative of \(\cos(x)\) is \(-\sin(x)\).
This technique helps in tackling complex problems that incorporate trigonometric terms, enabling a clearer and more organized approach to solving derivatives.
Chain Rule in Differentiation
The chain rule is a fundamental differentiation technique used to differentiate composite functions. It is especially useful when a function is nested within another function.
The chain rule states that if you have a function \(y = f(g(x))\), the derivative \(y'\) is found by multiplying the derivative of the outer function \(f\) by the derivative of the inner function \(g\): \( y' = f'(g(x)) \cdot g'(x)\).
Although the chain rule was mentioned in the original step-by-step solution, it wasn't directly needed in this particular problem because both terms were either simple polynomials or standard trigonometric functions.
Understanding the chain rule, however, is essential for more advanced calculus problems where functions are compounded, allowing you to break down complex tasks into manageable parts.
The chain rule states that if you have a function \(y = f(g(x))\), the derivative \(y'\) is found by multiplying the derivative of the outer function \(f\) by the derivative of the inner function \(g\): \( y' = f'(g(x)) \cdot g'(x)\).
Although the chain rule was mentioned in the original step-by-step solution, it wasn't directly needed in this particular problem because both terms were either simple polynomials or standard trigonometric functions.
Understanding the chain rule, however, is essential for more advanced calculus problems where functions are compounded, allowing you to break down complex tasks into manageable parts.
Other exercises in this chapter
Problem 1
Find the linearization \(L(x)\) of \(f(x)\) at \(x=a.\) \(f(x)=x^{3}-2 x+3, \quad a=2\)
View solution Problem 1
Use implicit differentiation to find \(d y / d x\). \(x^{2} y+x y^{2}=6\)
View solution Problem 1
Exercises \(1-6\) give the positions \(s=f(t)\) of a body moving on a coordinate line, with \(s\) in meters and \(t\) in seconds. a. Find the body's displacemen
View solution Problem 1
In Exercises \(1-12,\) find the first and second derivatives. \begin{equation} y=-x^{2}+3 \end{equation}
View solution