Find dy/dx y = square root of x(x-2)
The question is asking you to determine the derivative of the given function with respect to x, which means you need to find the rate of change of the function y with respect to changes in x. The function provided is y = √x(x-2), and you are required to apply differentiation rules to find dy/dx, which is the mathematical expression for the derivative of y with respect to x. This will involve using the chain rule and the product rule, as the function is a product of two terms, one of which is also a composite function (the square root of x).
$y = \sqrt{x} \left(\right. x - 2 \left.\right)$
Convert the square root into an exponent using the rule $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$. Thus, $\sqrt{x}$ becomes $x^{\frac{1}{2}}$. The equation now is $y = x^{\frac{1}{2}}(x - 2)$.
Take the derivative of both sides with respect to $x$: $\frac{d}{dx}(y) = \frac{d}{dx}\left(x^{\frac{1}{2}}(x - 2)\right)$.
The derivative of $y$ with respect to $x$ is denoted by $\frac{dy}{dx}$.
Apply the derivative to the right-hand side of the equation.
Use the Product Rule: $\frac{d}{dx}[f(x)g(x)] = f(x)\frac{d}{dx}[g(x)] + g(x)\frac{d}{dx}[f(x)]$, where $f(x) = x^{\frac{1}{2}}$ and $g(x) = x - 2$. This gives us $x^{\frac{1}{2}}\frac{d}{dx}[x - 2] + (x - 2)\frac{d}{dx}[x^{\frac{1}{2}}]$.
Proceed with differentiation.
Apply the Sum Rule: the derivative of $x - 2$ is the sum of the derivatives of its terms. We get $x^{\frac{1}{2}}(1 + 0) + (x - 2)\frac{d}{dx}[x^{\frac{1}{2}}]$.
Use the Power Rule: $\frac{d}{dx}[x^n] = nx^{n-1}$, where $n = 1$. This simplifies to $x^{\frac{1}{2}} + (x - 2)\frac{d}{dx}[x^{\frac{1}{2}}]$.
The derivative of a constant is zero, so the derivative of $-2$ is $0$. We now have $x^{\frac{1}{2}} + (x - 2)\frac{d}{dx}[x^{\frac{1}{2}}]$.
Simplify the expression.
Add $1$ and $0$ together to get $x^{\frac{1}{2}} + (x - 2)\frac{d}{dx}[x^{\frac{1}{2}}]$.
Multiplying $x^{\frac{1}{2}}$ by $1$ does not change its value.
Apply the Power Rule again with $n = \frac{1}{2}$ to get $x^{\frac{1}{2}} + (x - 2)\left(\frac{1}{2}x^{\frac{1}{2} - 1}\right)$.
To express $-1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$.
Combine the terms to get $x^{\frac{1}{2}} + (x - 2)\left(\frac{1}{2}x^{\frac{-1}{2}}\right)$.
Simplify the expression further.
Multiply $-1$ by $2$.
Subtract $2$ from $1$ to get $x^{\frac{1}{2}} + (x - 2)\left(\frac{1}{2}x^{\frac{-1}{2}}\right)$.
Move the negative exponent to the denominator using the rule $b^{-n} = \frac{1}{b^n}$.
Combine the fraction to get $x^{\frac{1}{2}} + (x - 2)\frac{1}{2x^{\frac{1}{2}}}$.
Simplify by distributing and combining like terms.
Factor out common terms and cancel where possible.
Reformulate the equation by setting the left side equal to the simplified right side: $y = \frac{3x^{\frac{1}{2}}}{2} - \frac{1}{x^{\frac{1}{2}}}$.
Replace $y$ with $\frac{dy}{dx}$ to get the final derivative: $\frac{dy}{dx} = \frac{3x^{\frac{1}{2}}}{2} - \frac{1}{x^{\frac{1}{2}}}$.
The Product Rule is used when differentiating products of two functions: $\frac{d}{dx}[f(x)g(x)] = f(x)\frac{d}{dx}[g(x)] + g(x)\frac{d}{dx}[f(x)]$.
The Sum Rule states that the derivative of a sum of functions is the sum of their derivatives.
The Power Rule for differentiation is given by $\frac{d}{dx}[x^n] = nx^{n-1}$.
A constant's derivative with respect to any variable is zero.
Negative exponents indicate reciprocals: $b^{-n} = \frac{1}{b^n}$.
Simplifying expressions often involves factoring, distributing, and combining like terms.
The derivative of a function represents the rate at which the function's value changes with respect to changes in the input value.