Find dy/dx x+y^3-xy=1
The question requests the calculation of the derivative of y with respect to x, commonly denoted as dy/dx, for the given implicit equation x + y^3 - xy = 1. To solve this, one would need to use implicit differentiation, which is a technique applied when a function cannot be easily solved for one variable in terms of another. In this particular equation, y is not explicitly solved in terms of x, so you must differentiate both sides of the equation with respect to x, treating y as a function of x (y(x)) and applying the chain rule where necessary.
$x + y^{3} - x y = 1$
Answer
Solution:
Step:1 Perform differentiation on each term of the equation with respect to $x$: $\frac{d}{dx}(x + y^3 - xy) = \frac{d}{dx}(1)$
Step:2 Apply differentiation to the left-hand side term by term.
Step:2.1 Begin differentiation.
Step:2.1.1 Utilize the Sum Rule in differentiation, which allows us to differentiate each term separately: $\frac{d}{dx}(x) + \frac{d}{dx}(y^3) + \frac{d}{dx}(-xy)$.
Step:2.1.2 Apply the Power Rule for differentiation to $x$, which states that the derivative of $x^n$ is $nx^{n-1}$, where $n=1$: $1 + \frac{d}{dx}(y^3) + \frac{d}{dx}(-xy)$.
Step:2.2 Compute the derivative of $y^3$ with respect to $x$.
Step:2.2.1 Employ the Chain Rule for differentiation, which involves taking the derivative of the outer function and multiplying it by the derivative of the inner function: $\frac{d}{dx}(f(g(x))) = f'(g(x))g'(x)$, where $f(x) = x^3$ and $g(x) = y$.
Step:2.2.1.1 Introduce a substitution $u = y$ to apply the Chain Rule: $1 + \frac{d}{du}(u^3)\frac{d}{dx}(y) + \frac{d}{dx}(-xy)$.
Step:2.2.1.2 Differentiate $u^3$ using the Power Rule, where $n=3$: $1 + 3u^2\frac{d}{dx}(y) + \frac{d}{dx}(-xy)$.
Step:2.2.1.3 Replace $u$ back with $y$: $1 + 3y^2\frac{d}{dx}(y) + \frac{d}{dx}(-xy)$.
Step:2.2.2 Express $\frac{d}{dx}(y)$ as $\frac{dy}{dx}$: $1 + 3y^2\frac{dy}{dx} + \frac{d}{dx}(-xy)$.
Step:2.3 Determine the derivative of $-xy$ with respect to $x$.
Step:2.3.1 Recognize that $-1$ is a constant factor and apply differentiation to $-xy$: $1 + 3y^2\frac{dy}{dx} - \frac{d}{dx}(xy)$.
Step:2.3.2 Apply the Product Rule for differentiation, which states that the derivative of $f(x)g(x)$ is $f(x)g'(x) + g(x)f'(x)$, where $f(x) = x$ and $g(x) = y$: $1 + 3y^2\frac{dy}{dx} - (x\frac{d}{dx}(y) + y\frac{d}{dx}(x))$.
Step:2.3.3 Express $\frac{d}{dx}(y)$ as $\frac{dy}{dx}$: $1 + 3y^2\frac{dy}{dx} - (xy + y\frac{d}{dx}(x))$.
Step:2.3.4 Differentiate $x$ using the Power Rule, where $n=1$: $1 + 3y^2\frac{dy}{dx} - (xy + y \cdot 1)$.
Step:2.3.5 Simplify the multiplication of $y$ by $1$: $1 + 3y^2\frac{dy}{dx} - (xy + y)$.
Step:2.4 Simplify the expression.
Step:2.4.1 Apply the distributive property to expand the terms: $1 + 3y^2\frac{dy}{dx} - xy - y$.
Step:2.4.2 Remove any unnecessary parentheses: $1 + 3y^2\frac{dy}{dx} - xy - y$.
Step:2.4.3 Reorder the terms for clarity: $3y^2\frac{dy}{dx} - xy - y + 1$.
Step:3 Differentiate the constant $1$ on the right-hand side, which yields $0$: $0$.
Step:4 Combine the derivatives from both sides to form an equation: $3y^2\frac{dy}{dx} - xy - y + 1 = 0$.
Step:5 Isolate $\frac{dy}{dx}$ to solve for the derivative.
Step:5.1 Move all terms not containing $\frac{dy}{dx}$ to the other side of the equation.
Step:5.1.1 Add $xy$ and $y$ to both sides: $3y^2\frac{dy}{dx} = xy + y + 1$.
Step:5.1.2 Subtract $1$ from both sides to isolate terms with $\frac{dy}{dx}$: $3y^2\frac{dy}{dx} = xy + y$.
Step:5.2 Extract $\frac{dy}{dx}$ from the left side.
Step:5.2.1 Divide both sides by $3y^2$: $\frac{dy}{dx} = \frac{xy + y}{3y^2}$.
Step:6 Finally, express $\frac{dy}{dx}$ in its simplest form: $\frac{dy}{dx} = \frac{x + 1}{3y}$.
Knowledge Notes:
The problem involves finding the derivative of $y$ with respect to $x$ from the given implicit equation $x + y^3 - xy = 1$. The solution employs several rules of differentiation:
Sum Rule: The derivative of a sum of functions is the sum of their derivatives.
Power Rule: The derivative of $x^n$ with respect to $x$ is $nx^{n-1}$.
Chain Rule: The derivative of a composite function $f(g(x))$ is $f'(g(x))g'(x)$.
Product Rule: The derivative of a product of two functions $f(x)g(x)$ is $f(x)g'(x) + g(x)f'(x)$.
The process involves differentiating each term of the equation separately, applying the appropriate rule, and then combining the results to solve for $\frac{dy}{dx}$. The final step is to isolate $\frac{dy}{dx}$ and simplify the expression to find the derivative of $y$ with respect to $x$.
