Problem 29
Question
Find the derivatives of all orders of the functions $$ y=\frac{x^{4}}{2}-\frac{3}{2} x^{2}-x $$
Step-by-Step Solution
Verified Answer
The first four derivatives are \(y' = 2x^3 - 3x - 1\), \(y'' = 6x^2 - 3\), \(y''' = 12x\), and \(y^{(4)} = 12\); for \(n \geq 5\), \(y^{(n)} = 0\).
1Step 1: Identify the Function and Its Components
The given function is \(y = \frac{x^4}{2} - \frac{3}{2}x^2 - x\). This is a polynomial function consisting of three terms: \(\frac{x^4}{2}\), \(-\frac{3}{2}x^2\), and \(-x\). Each term needs to be differentiated separately.
2Step 2: Differentiate the Function to Find the First Derivative
Apply the power rule of differentiation, which states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). - The derivative of \(\frac{x^4}{2}\) is \(\frac{4}{2}x^{4-1} = 2x^3\).- The derivative of \(-\frac{3}{2}x^2\) is \(-\frac{3}{2} \cdot 2x^{2-1} = -3x\).- The derivative of \(-x\) is \(-1\).Combine these to get the first derivative: \(y' = 2x^3 - 3x - 1\).
3Step 3: Find the Second Derivative
Differentiate the first derivative \(y' = 2x^3 - 3x - 1\).- The derivative of \(2x^3\) is \(6x^2\).- The derivative of \(-3x\) is \(-3\).- The derivative of \(-1\) is \(0\).The second derivative is \(y'' = 6x^2 - 3\).
4Step 4: Find the Third Derivative
Differentiate the second derivative \(y'' = 6x^2 - 3\).- The derivative of \(6x^2\) is \(12x\).- The derivative of \(-3\) is \(0\).The third derivative is \(y''' = 12x\).
5Step 5: Find the Fourth Derivative
Differentiate the third derivative \(y''' = 12x\).- The derivative of \(12x\) is \(12\).The fourth derivative is \(y^{(4)} = 12\).
6Step 6: Higher Order Derivatives
For derivatives of order higher than four, since the function simplifies to a constant (12), any higher derivative will be zero. Hence, \(y^{(n)} = 0\) for all \(n \geq 5\).
Key Concepts
Power Rule of DifferentiationHigher Order DerivativesPolynomial Differentiation
Power Rule of Differentiation
The power rule of differentiation is fundamental when working with polynomial functions. It allows us to find the derivative of each term individually. The rule states that for a term of the form \(a \cdot x^n\), the derivative is \(n \cdot a \cdot x^{n-1}\). Simply put:
- Multiply the exponent \(n\) by the coefficient \(a\).
- Decrease the exponent by one.
Higher Order Derivatives
Higher order derivatives involve taking the derivative of a function multiple times. Using our exercise as an example, we don't just stop at the first derivative. We also need to find the second, third, and so forth:
- The first derivative \(y'\) provides the rate of change of the function \(y\).
- The second derivative \(y''\) can help identify concavity and inflection points, and is the derivative of \(y'\).
- The third derivative \(y'''\) continues this pattern and often relates to the rate of change of acceleration.
- As you move to the fourth derivative and beyond, your polynomial terms might reduce to constants, as in our exercise where the fourth derivative \(y^{(4)}\) equals 12, and any higher derivative results in zero.
Polynomial Differentiation
Polynomial differentiation is about finding derivatives for functions formed by sums of monomial terms like \(ax^n\). Each term in a polynomial behaves independently when taking derivatives:
Polynomial differentiation is not only essential in calculus, but also lays groundwork for understanding and analyzing complex systems in various fields such as economics, biology, and data sciences.
- Determine each term’s derivative separately using the power rule.
- Reassemble the components to gain the overall derivative.
Polynomial differentiation is not only essential in calculus, but also lays groundwork for understanding and analyzing complex systems in various fields such as economics, biology, and data sciences.
Other exercises in this chapter
Problem 29
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In Exercises \(19-30,\) find \(d y\) $$ y=2 \cot \left(\frac{1}{\sqrt{x}}\right) $$
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Use implicit differentiation to find \(d y / d x\) in Exercises \(19-32\) $$ x+\sin y=x y $$
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