Answer :
To find the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] for the function:
[tex]\[ y = \sqrt[3]{\frac{x(x-2)}{x^2 + 1}} \][/tex]
we will use logarithmic differentiation. Here is a detailed step-by-step solution:
1. Take the Natural Logarithm of Both Sides:
Since [tex]\( y = \left(\frac{x(x-2)}{x^2 + 1}\right)^{1/3} \)[/tex], we can take the natural logarithm of both sides:
[tex]\[ \ln y = \ln \left(\left(\frac{x(x-2)}{x^2 + 1}\right)^{1/3}\right) \][/tex]
2. Simplify Using Logarithm Properties:
Using the property of logarithms that [tex]\(\ln\left(a^b\right) = b \ln(a)\)[/tex], we get:
[tex]\[ \ln y = \frac{1}{3} \ln \left(\frac{x(x-2)}{x^2 + 1}\right) \][/tex]
3. Break the Logarithm of a Quotient:
Using the property that [tex]\(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)[/tex], we get:
[tex]\[ \ln y = \frac{1}{3} \left( \ln (x(x-2)) - \ln(x^2 + 1) \right) \][/tex]
4. Break the Logarithm of a Product:
Using the property that [tex]\(\ln(ab) = \ln(a) + \ln(b)\)[/tex], we get:
[tex]\[ \ln (x(x-2)) = \ln x + \ln (x-2) \][/tex]
Therefore:
[tex]\[ \ln y = \frac{1}{3} \left( \ln x + \ln (x-2) - \ln(x^2 + 1) \right) \][/tex]
5. Differentiate Both Sides with Respect to [tex]\( x \)[/tex]:
Since [tex]\(\ln y\)[/tex] is a function of [tex]\( y(x) \)[/tex], we can differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} \left( \ln y \right) = \frac{d}{dx} \left( \frac{1}{3} (\ln x + \ln (x-2) - \ln(x^2 + 1)) \right) \][/tex]
6. Apply the Chain Rule:
The derivative of [tex]\(\ln y\)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\(\frac{1}{y} \frac{dy}{dx}\)[/tex]. Therefore:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \left( \frac{d}{dx} (\ln x) + \frac{d}{dx} (\ln (x-2)) - \frac{d}{dx} (\ln (x^2 + 1)) \right) \][/tex]
7. Compute the Derivatives:
[tex]\[ \frac{d}{dx} (\ln x) = \frac{1}{x}, \quad \frac{d}{dx} (\ln (x-2)) = \frac{1}{x-2}, \quad \text{and} \quad \frac{d}{dx} (\ln (x^2 + 1)) = \frac{2x}{x^2 + 1} \][/tex]
Therefore:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2 + 1} \right) \][/tex]
8. Combine the Terms:
Multiply both sides by [tex]\( y \)[/tex]:
[tex]\[ \frac{dy}{dx} = y \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2 + 1} \right) \][/tex]
Recall [tex]\( y = \left( \frac{x(x-2)}{x^2 + 1} \right)^{1/3} \)[/tex]:
[tex]\[ \frac{dy}{dx} = \left( \frac{x(x-2)}{x^2 + 1} \right)^{1/3} \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2 + 1} \right) \][/tex]
We have obtained the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dy}{dx} = \left( \frac{x(x-2)}{x^2 + 1} \right)^{1/3} \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2 + 1} \right) \][/tex]
To summarize, we used the technique of logarithmic differentiation to find:
[tex]\[ \frac{dy}{dx} = \left( \frac{x(x-2)}{x^2 + 1} \right)^{1/3} \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2 + 1} \right) \][/tex]
[tex]\[ y = \sqrt[3]{\frac{x(x-2)}{x^2 + 1}} \][/tex]
we will use logarithmic differentiation. Here is a detailed step-by-step solution:
1. Take the Natural Logarithm of Both Sides:
Since [tex]\( y = \left(\frac{x(x-2)}{x^2 + 1}\right)^{1/3} \)[/tex], we can take the natural logarithm of both sides:
[tex]\[ \ln y = \ln \left(\left(\frac{x(x-2)}{x^2 + 1}\right)^{1/3}\right) \][/tex]
2. Simplify Using Logarithm Properties:
Using the property of logarithms that [tex]\(\ln\left(a^b\right) = b \ln(a)\)[/tex], we get:
[tex]\[ \ln y = \frac{1}{3} \ln \left(\frac{x(x-2)}{x^2 + 1}\right) \][/tex]
3. Break the Logarithm of a Quotient:
Using the property that [tex]\(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)[/tex], we get:
[tex]\[ \ln y = \frac{1}{3} \left( \ln (x(x-2)) - \ln(x^2 + 1) \right) \][/tex]
4. Break the Logarithm of a Product:
Using the property that [tex]\(\ln(ab) = \ln(a) + \ln(b)\)[/tex], we get:
[tex]\[ \ln (x(x-2)) = \ln x + \ln (x-2) \][/tex]
Therefore:
[tex]\[ \ln y = \frac{1}{3} \left( \ln x + \ln (x-2) - \ln(x^2 + 1) \right) \][/tex]
5. Differentiate Both Sides with Respect to [tex]\( x \)[/tex]:
Since [tex]\(\ln y\)[/tex] is a function of [tex]\( y(x) \)[/tex], we can differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} \left( \ln y \right) = \frac{d}{dx} \left( \frac{1}{3} (\ln x + \ln (x-2) - \ln(x^2 + 1)) \right) \][/tex]
6. Apply the Chain Rule:
The derivative of [tex]\(\ln y\)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\(\frac{1}{y} \frac{dy}{dx}\)[/tex]. Therefore:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \left( \frac{d}{dx} (\ln x) + \frac{d}{dx} (\ln (x-2)) - \frac{d}{dx} (\ln (x^2 + 1)) \right) \][/tex]
7. Compute the Derivatives:
[tex]\[ \frac{d}{dx} (\ln x) = \frac{1}{x}, \quad \frac{d}{dx} (\ln (x-2)) = \frac{1}{x-2}, \quad \text{and} \quad \frac{d}{dx} (\ln (x^2 + 1)) = \frac{2x}{x^2 + 1} \][/tex]
Therefore:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2 + 1} \right) \][/tex]
8. Combine the Terms:
Multiply both sides by [tex]\( y \)[/tex]:
[tex]\[ \frac{dy}{dx} = y \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2 + 1} \right) \][/tex]
Recall [tex]\( y = \left( \frac{x(x-2)}{x^2 + 1} \right)^{1/3} \)[/tex]:
[tex]\[ \frac{dy}{dx} = \left( \frac{x(x-2)}{x^2 + 1} \right)^{1/3} \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2 + 1} \right) \][/tex]
We have obtained the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dy}{dx} = \left( \frac{x(x-2)}{x^2 + 1} \right)^{1/3} \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2 + 1} \right) \][/tex]
To summarize, we used the technique of logarithmic differentiation to find:
[tex]\[ \frac{dy}{dx} = \left( \frac{x(x-2)}{x^2 + 1} \right)^{1/3} \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2 + 1} \right) \][/tex]
