Chapter 23: Problem 13
Find the derivative of each function. Check some by calculator. $$y=\sqrt{1-3 x^{2}}$$
Short Answer
Expert verified
\(y' = -3x(1-3x^2)^{-\frac{1}{2}}\)
Step by step solution
01
Identify the function
First, recognize the function that needs differentiation, which is the square root of a binomial: \(y=\sqrt{1-3x^2}\).
02
Rewrite the function
To simplify the differentiation process, rewrite the square root as a power: \(y=(1-3x^2)^{\frac{1}{2}}\).
03
Apply the chain rule
Use the chain rule, which states that the derivative of \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\). Here \(f(u) = u^{\frac{1}{2}}\) and \(g(x) = 1-3x^2\).
04
Differentiate the outer function
Differentiate \(f(u)\) with respect to \(u\): \(f'(u) = \frac{1}{2}u^{-\frac{1}{2}}\).
05
Differentiate the inner function
Differentiate \(g(x)\) with respect to \(x\): \(g'(x) = -3(2x) = -6x\).
06
Combine derivatives using the chain rule
Multiply \(f'(u)\) by \(g'(x)\), substituting \(u\) with \(g(x)\): \(y' = \frac{1}{2}(1-3x^2)^{-\frac{1}{2}} \cdot (-6x)\).
07
Simplify the derivative
Simplify the expression to get the final derivative: \(y' = -3x(1-3x^2)^{-\frac{1}{2}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
Understanding the chain rule is essential when working with complex functions, especially those involving compositions of multiple functions. The chain rule is a powerful tool in calculus that provides us with a method to differentiate composite functions, which are functions that exist within other functions.
For example, if you have a function in the form of \( f(g(x)) \), the chain rule states the derivative is found using \( f'(g(x)) \cdot g'(x) \). This means that we take the derivative of the outer function, \( f \), with respect to the inner function, \( g \), and then multiply it by the derivative of the inner function, \( g \), with respect to \( x \).
Applied to the given exercise, with the square root function \( y = \sqrt{1-3x^2} \), we recognized the need to differentiate not just the square root (the outer function) but the binomial expression \( 1-3x^2 \) (the inner function) as well. By following the chain rule steps meticulously, we calculated the derivative with precision, crucial for functions where direct substitution is not readily available.
For example, if you have a function in the form of \( f(g(x)) \), the chain rule states the derivative is found using \( f'(g(x)) \cdot g'(x) \). This means that we take the derivative of the outer function, \( f \), with respect to the inner function, \( g \), and then multiply it by the derivative of the inner function, \( g \), with respect to \( x \).
Applied to the given exercise, with the square root function \( y = \sqrt{1-3x^2} \), we recognized the need to differentiate not just the square root (the outer function) but the binomial expression \( 1-3x^2 \) (the inner function) as well. By following the chain rule steps meticulously, we calculated the derivative with precision, crucial for functions where direct substitution is not readily available.
Power Rule of Derivatives
The power rule is another fundamental concept in differentiation that dramatically simplifies the process of finding derivatives, especially when dealing with monomials—single-term algebraic expressions—and their powers.
The power rule simply states that if you have a function \( f(x) = x^n \), where \( n \) is any real number, the derivative of \( f \) with respect to \( x \) is \( f'(x) = nx^{n-1} \).
In our exercise, we used this rule when we rewrote the square root function as a power, transforming \( y = \sqrt{1-3x^2} \) into \( y = (1-3x^2)^{\frac{1}{2}} \), and then we differentiated \( f(u) = u^{\frac{1}{2}} \) using the power rule to get \( f'(u) = \frac{1}{2}u^{-\frac{1}{2}} \). This transformation and subsequent application of the power rule are imperative for handling square roots and other rational exponents successfully in differentiation.
The power rule simply states that if you have a function \( f(x) = x^n \), where \( n \) is any real number, the derivative of \( f \) with respect to \( x \) is \( f'(x) = nx^{n-1} \).
In our exercise, we used this rule when we rewrote the square root function as a power, transforming \( y = \sqrt{1-3x^2} \) into \( y = (1-3x^2)^{\frac{1}{2}} \), and then we differentiated \( f(u) = u^{\frac{1}{2}} \) using the power rule to get \( f'(u) = \frac{1}{2}u^{-\frac{1}{2}} \). This transformation and subsequent application of the power rule are imperative for handling square roots and other rational exponents successfully in differentiation.
Derivatives of Binomial Expressions
Binomial expressions consist of two terms separated by a plus or minus sign. When we differentiate a function that contains a binomial expression, we treat each term separately according to the rules of differentiation.
In this exercise, the binomial expression in question was \( 1-3x^2 \). Here, the constant, '1', simply vanishes upon differentiation, reflecting the fact that the derivative of a constant is zero. Conversely, the term \( -3x^2 \) is differentiated using a combination of the power rule and the constant multiple rule. The constant multiple rule states that the derivative of a constant times a function is just the constant times the derivative of the function, which led us to \( g'(x) = -3(2x) \) or simplified to \( -6x \).
By combining these rules, we can tackle the differentiation of complex expressions one step at a time, building up to a full derivative by addressing each piece individually.
In this exercise, the binomial expression in question was \( 1-3x^2 \). Here, the constant, '1', simply vanishes upon differentiation, reflecting the fact that the derivative of a constant is zero. Conversely, the term \( -3x^2 \) is differentiated using a combination of the power rule and the constant multiple rule. The constant multiple rule states that the derivative of a constant times a function is just the constant times the derivative of the function, which led us to \( g'(x) = -3(2x) \) or simplified to \( -6x \).
By combining these rules, we can tackle the differentiation of complex expressions one step at a time, building up to a full derivative by addressing each piece individually.
