Find dy/dx x+y^3-xy=1

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 $n{x}^{n-1}$, 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: $\frac{d}{dx}(f(g(x)))=f^{\prime }(g(x))g^{\prime }(x)$, 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 $f(x)g^{\prime }(x)+g(x)f^{\prime }(x)$, 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:

1. Sum Rule: The derivative of a sum of functions is the sum of their derivatives.

2. Power Rule: The derivative of $x^n$ with respect to $x$ is $n{x}^{n-1}$.

3. Chain Rule: The derivative of a composite function $f(g(x))$ is $f^{\prime }(g(x))g^{\prime }(x)$.

4. Product Rule: The derivative of a product of two functions $f(x)g(x)$ is $f(x)g^{\prime }(x)+g(x)f^{\prime }(x)$.

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 $x$.