Problem 1
Question
Differentiate the functions given in Problems 1-22 with respect to the independent variable. $$ f(x)=4 x^{3}-7 x+1 $$
Step-by-Step Solution
Verified Answer
The derivative of the function \( f(x) = 4x^3 - 7x + 1 \) is \( 12x^2 - 7 \).
1Step 1: Differentiate term by term
To differentiate the function, differentiate each term separately based on differentiation rules. The function is given by \( f(x) = 4x^3 - 7x + 1 \).
2Step 2: Differentiate the first term
The first term is \( 4x^3 \). Use the power rule \( \frac{d}{dx}[x^n] = nx^{n-1} \) to differentiate: \[ \frac{d}{dx}[4x^3] = 4 \cdot 3x^{3-1} = 12x^2 \]
3Step 3: Differentiate the second term
The second term is \( -7x \). Again, use the power rule where the exponent is 1: \[ \frac{d}{dx} [-7x] = -7 \cdot 1x^{1-1} = -7 \]
4Step 4: Differentiate the third term
The third term is a constant \( 1 \). The derivative of any constant is 0: \[ \frac{d}{dx}[1] = 0 \]
5Step 5: Combine the derivatives of each term
Summing the derivatives from each term gives the overall derivative of the function: \[ \frac{d}{dx}[f(x)] = 12x^2 - 7 + 0 = 12x^2 - 7 \]
Key Concepts
Power RuleDerivativesConstant Term Differentiation
Power Rule
Understanding differentiation can be quite exciting, especially with a simple rule like the Power Rule. This rule states that if you have a function of the form \( x^n \), then its derivative is given by \( nx^{n-1} \). Here's why it's handy:
- Applies to any function where a variable is raised to a power.
- Allows you to quickly find the rate of change of that function.
- Easy to use and remember, even if the exponent \( n \) is a fraction or a negative number.
Derivatives
Derivatives play a fundamental role in calculus and help us understand how a function changes. Think of a derivative as a tool that measures the sensitivity of a function's output to changes in its input. It's especially useful when:
- Determining the slope of a function at any given point.
- Finding the rate at which something changes, such as speed or growth.
- Identifying maximum and minimum points of a function, also known as critical points.
Constant Term Differentiation
Constant term differentiation is perhaps the simplest rule when working with derivatives. It says that the derivative of any constant is always zero. Why? Because constants do not change, and thus their rate of change is zero. In simple terms:
- A constant remains fixed, irrespective of the variable changes.
- This makes constant terms disappear in the derivative expression since their contribution to change is null.
- Simplifies calculations when dealing with polynomials or functions combining multiple terms.
Other exercises in this chapter
Problem 1
Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ f(x)=e^{3 x} $$
View solution Problem 1
In Problems 1-16, use the product rule to find the derivative with respect to the independent variable. $$ f(x)=(x+5)\left(x^{2}-3\right) $$
View solution Problem 1
Find the inverse of each function and differentiate each inverse in two ways: (i) Differentiate the inverse function directly, and (ii) use (4.12) to find the d
View solution Problem 2
Use the formula $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ to approximate the value of the given function. Then compare your result with the value you get from a
View solution