Find dy/dx 3x^2-2xy+3y=1
The question provided is asking for the derivative of the equation 3x^2 - 2xy + 3y = 1 with respect to x, commonly denoted as dy/dx. In calculus, finding dy/dx of an equation involving both x and y variables typically involves differentiating implicitly, because y is considered a function of x (y = f(x)) even though it is not solved for explicitly. The process will require the use of implicit differentiation rules, which take the derivatives of both sides of the equation with respect to x while applying the chain rule to the terms involving y since y is a function of x.
$3 x^{2} - 2 x y + 3 y = 1$
Take the derivative of both sides of the equation with respect to $x$: $\frac{d}{dx}(3x^2 - 2xy + 3y) = \frac{d}{dx}(1)$.
Differentiate each term on the left side separately.
Apply the Sum Rule in differentiation: $\frac{d}{dx}(3x^2) + \frac{d}{dx}(-2xy) + \frac{d}{dx}(3y)$.
Differentiate $3x^2$ with respect to $x$.
The constant $3$ remains, and the derivative of $x^2$ is $2x$, resulting in $3(2x)$.
Simplify to get $6x$.
Differentiate $-2xy$ with respect to $x$.
The constant $-2$ remains, and apply the Product Rule to $xy$: $-2(x\frac{d}{dx}(y) + y\frac{d}{dx}(x))$.
Recognize that $\frac{d}{dx}(y)$ is $\frac{dy}{dx}$ and $\frac{d}{dx}(x)$ is $1$.
Simplify to $-2(xy' + y)$.
Differentiate $3y$ with respect to $x$.
The constant $3$ remains, and the derivative of $y$ is $\frac{dy}{dx}$, resulting in $3\frac{dy}{dx}$.
Combine all the differentiated terms.
Combine like terms to simplify the expression: $6x - 2(xy' + y) + 3y'$.
Differentiate the right side of the equation, which is a constant, to get $0$.
Set the left side equal to the right side to form the equation: $6x - 2(xy' + y) + 3y' = 0$.
Solve for $\frac{dy}{dx}$.
Isolate terms involving $\frac{dy}{dx}$ on one side and the rest on the other side.
Move $6x$ to the right side: $-2xy' - 2y + 3y' = -6x$.
Combine $-2y$ and $3y'$: $-2xy' + y' = -6x + 2y$.
Factor out $\frac{dy}{dx}$ from the left side.
Factor $\frac{dy}{dx}$ out of $-2xy'$ and $y'$: $\frac{dy}{dx}(-2x + 1) = -6x + 2y$.
Divide both sides by $-2x + 1$ to solve for $\frac{dy}{dx}$.
Divide and simplify: $\frac{dy}{dx} = \frac{-6x + 2y}{-2x + 1}$.
Sum Rule: The derivative of a sum of functions is the sum of the derivatives.
Constant Multiple Rule: The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function.
Power Rule: The derivative of $x^n$ with respect to $x$ is $nx^{n-1}$.
Product Rule: The derivative of a product of two functions $f(x)g(x)$ is $f'(x)g(x) + f(x)g'(x)$.
Chain Rule: The derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
Implicit Differentiation: When a function is not given explicitly as $y=f(x)$, but implicitly as a relation between $x$ and $y$, we differentiate both sides of the equation with respect to $x$ and solve for $\frac{dy}{dx}$.
Derivative of a Constant: The derivative of a constant is zero.
Derivative of $y$ with respect to $x$: When differentiating $y$ with respect to $x$, we denote it as $\frac{dy}{dx}$, which represents the rate of change of $y$ with respect to $x$.