Problem 4
Question
Find the derivatives of the functions in exercise. Even if \(n\) is negative or a fraction, the derivative of \(x^{n}\) is \(n x^{n-1}\). $$ 1+(7 / x)+\left(5 / x^{2}\right) $$
Step-by-Step Solution
Verified Answer
The derivative is \(-7x^{-2} - 10x^{-3}\).
1Step 1: Simplify to a Polynomial
Start by rewriting the expression so that it represents a polynomial. Specifically, we'll rewrite the terms with negative exponents: \(7/x\) becomes \(7x^{-1}\) and \(5/x^2\) becomes \(5x^{-2}\). The expression now is \(1 + 7x^{-1} + 5x^{-2}\).
2Step 2: Apply the Power Rule
For each term in the polynomial, apply the derivative power rule, \(\frac{d}{dx} x^n = nx^{n-1}\). - For the constant \(1\), the derivative is \(0\).- For the term \(7x^{-1}\), the derivative is \(-7x^{-2}\) (since \(n=-1\)). - For the term \(5x^{-2}\), the derivative is \(-10x^{-3}\) (since \(n=-2\) and multiplying by the exponent gives \(-10\)).
3Step 3: Combine the Results
Combine the derivatives of each term to get the final expression. The derivative of the entire expression is \(-7x^{-2} - 10x^{-3}\).
Key Concepts
Understanding the Power RuleNegative Exponents in DifferentiationFractional Exponents Explained
Understanding the Power Rule
The power rule is one of the most fundamental tools in calculus for finding the derivative of a function. It's especially handy when dealing with polynomials, as it simplifies the process of differentiation significantly. The rule states that if you have a function of the form \(x^n\), where \(n\) can be any real number, its derivative is \(nx^{n-1}\).
Here's a simple breakdown of how the power rule works:- Identify the exponent \(n\) of the function \(x^n\). - Multiply this exponent by the coefficient of \(x\). - Subtract one from the exponent. For example, differentiating \(x^3\) requires us to multiply the exponent 3 by the coefficient (which is 1 in this case) and then reduce the exponent by 1, resulting in \(3x^2\). The power rule makes differentiating complex terms much straightforward, even when dealing with higher powers.
It's important to note that the power rule holds true even when \(n\) is negative or a fraction, which brings us to our next concept.
Here's a simple breakdown of how the power rule works:- Identify the exponent \(n\) of the function \(x^n\). - Multiply this exponent by the coefficient of \(x\). - Subtract one from the exponent. For example, differentiating \(x^3\) requires us to multiply the exponent 3 by the coefficient (which is 1 in this case) and then reduce the exponent by 1, resulting in \(3x^2\). The power rule makes differentiating complex terms much straightforward, even when dealing with higher powers.
It's important to note that the power rule holds true even when \(n\) is negative or a fraction, which brings us to our next concept.
Negative Exponents in Differentiation
When you encounter negative exponents in a function, applying the power rule can seem a bit tricky, but it's actually quite easy. A negative exponent simply indicates reciprocal behavior. For example, \(x^{-1}\) is equivalent to \(1/x\). This transformation helps to rewrite terms in a form that's ready for differentiation.
Consider the function \(7x^{-1}\). To differentiate this term, you use the power rule by multiplying the exponent \(-1\) with the coefficient \(7\) to get \(-7\). Then, decrease the exponent by 1, resulting in the new term \(x^{-2}\). Hence, the derivative is \(-7x^{-2}\).
Another example is \(5x^{-2}\). Applying the power rule, multiply \(-2\) by the coefficient 5, giving \(-10\), and then subtract 1 from the exponent to get \(-3\). The resulting derivative is thus \(-10x^{-3}\).
Negative exponents can initially seem daunting, but through methodical application of the power rule, they transform swiftly into simpler expressions.
Consider the function \(7x^{-1}\). To differentiate this term, you use the power rule by multiplying the exponent \(-1\) with the coefficient \(7\) to get \(-7\). Then, decrease the exponent by 1, resulting in the new term \(x^{-2}\). Hence, the derivative is \(-7x^{-2}\).
Another example is \(5x^{-2}\). Applying the power rule, multiply \(-2\) by the coefficient 5, giving \(-10\), and then subtract 1 from the exponent to get \(-3\). The resulting derivative is thus \(-10x^{-3}\).
Negative exponents can initially seem daunting, but through methodical application of the power rule, they transform swiftly into simpler expressions.
Fractional Exponents Explained
Fractional exponents are another interesting aspect of differentiation you might encounter. These exponents denote roots and offer a unique approach to simplifying differentiation. For instance, expressing square roots and cube roots using fractional exponents makes applying the power rule a breeze.
For example, the square root of \(x\), written as \(\sqrt{x}\), can be expressed as \(x^{1/2}\). The power rule can then be applied by multiplying the exponent \(1/2\) by the coefficient (which is typically 1) and subtracting 1 from the exponent. This results in the derivative \((1/2)x^{-1/2}\). This expression simplifies to \(\frac{1}{2\sqrt{x}}\), keeping the mathematics seamless and simple.
Similarly, a cube root, such as \(\sqrt[3]{x}\), is \(x^{1/3}\). Following the same steps as before, the derivative becomes \((1/3)x^{-2/3}\).
The fractional exponent allows us to align root functions with the systematic rules of differentiation, making operations such as these more intuitive and efficient.
For example, the square root of \(x\), written as \(\sqrt{x}\), can be expressed as \(x^{1/2}\). The power rule can then be applied by multiplying the exponent \(1/2\) by the coefficient (which is typically 1) and subtracting 1 from the exponent. This results in the derivative \((1/2)x^{-1/2}\). This expression simplifies to \(\frac{1}{2\sqrt{x}}\), keeping the mathematics seamless and simple.
Similarly, a cube root, such as \(\sqrt[3]{x}\), is \(x^{1/3}\). Following the same steps as before, the derivative becomes \((1/3)x^{-2/3}\).
The fractional exponent allows us to align root functions with the systematic rules of differentiation, making operations such as these more intuitive and efficient.
Other exercises in this chapter
Problem 3
For \(f(x)=3 x\) and \(g(x)=1+3 x\), find \(f(4+h)\) and \(g(4+h)\) and \(f^{\prime}(4)\) and \(g^{\prime}(4)\). Sketch the graphs of \(f\) and \(g-\) why do th
View solution Problem 4
Decide whether \(A \Rightarrow B\) or \(B \Rightarrow A\) or neither or both: (a) \(A=\left[a_{n} \rightarrow 1\right] \quad B=\left[-a_{n} \rightarrow-1\right]
View solution Problem 4
The tangent line to \(y=x^{3}+6 x\) at the origin is \(y=\) Does it cross the curve again?
View solution Problem 5
If the sequence \(a_{1}, a_{2}, a_{3}, \ldots\) approaches zero, prove that we can put those numbers in any order and the new sequence still approaches zero.
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