Find dy/dx y^2+6y+9x=0

Solution:

Step 1:

Take the derivative of each term in the equation $y^2+6y+9x=0$ with respect to $x$.

$$\frac{d}{dx}(y^2+6y+9x)=\frac{d}{dx}(0)$$

Step 2:

Apply the derivative to each term separately.

Step 2.1:

Utilize the Sum Rule to differentiate the sum term by term.

$$\frac{d}{dx}(y^2)+\frac{d}{dx}(6y)+\frac{d}{dx}(9x)$$

Step 2.2:

Find the derivative of $y^2$ with respect to $x$.

Step 2.2.1:

Employ the Chain Rule for differentiation, where $f(x)=x^2$ and $g(x)=y$.

$$\frac{d}{du}(u^2)\frac{dy}{dx}+\frac{d}{dx}(6y)+\frac{d}{dx}(9x)$$

Step 2.2.2:

Apply the Power Rule, where the derivative of $u^n$ is $n{u}^{n-1}$, with $n=2$.

$$2u\frac{dy}{dx}+\frac{d}{dx}(6y)+\frac{d}{dx}(9x)$$

Step 2.2.3:

Substitute $u$ back with $y$.

$$2y\frac{dy}{dx}+\frac{d}{dx}(6y)+\frac{d}{dx}(9x)$$

Step 2.3:

Calculate the derivative of $6y$ with respect to $x$.

Step 2.3.1:

Since 6 is a constant, the derivative is $6\frac{dy}{dx}$.

$$2yy+6\frac{dy}{dx}+\frac{d}{dx}(9x)$$

Step 2.4:

Determine the derivative of $9x$ with respect to $x$.

Step 2.4.1:

As 9 is a constant, use the constant multiple rule to get $9\frac{dx}{dx}$.

$$2yy+6y+9\frac{dx}{dx}$$

Step 2.4.2:

Apply the Power Rule to $x^1$, yielding $1x^{1-1}$.

$$2yy+6y+9\cdot 1$$

Step 2.4.3:

Multiply 9 by 1 to simplify.

$$2yy+6y+9$$

Step 3:

The derivative of a constant, 0, is 0.

$$0$$

Step 4:

Combine the derivatives to form the equation.

$$2yy+6y+9=0$$

Step 5:

Isolate $y$ to solve for $\frac{dy}{dx}$.

Step 5.1:

Subtract 9 from both sides.

$$2yy+6y=-9$$

Step 5.2:

Factor out $2y$ from the terms on the left side.

Step 5.2.1:

Factor $2y$ from $2yy$.

$$2yy+6y=-9$$

Step 5.2.2:

Factor $2y$ from $6y$.

$$2yy+2y\cdot 3=-9$$

Step 5.2.3:

Extract $2y$ from the sum.

$$2y(y+3)=-9$$

Step 5.3:

Divide by $2(y+3)$ and simplify.

Step 5.3.1:

Divide each term by $2(y+3)$.

$$\frac{2y(y+3)}{2(y+3)}=\frac{-9}{2(y+3)}$$

Step 5.3.2:

Simplify the left side by canceling out common factors.

Step 5.3.2.1:

Cancel the 2.

$$\frac{\cancel{2}y(y+3)}{\cancel{2}(y+3)}=\frac{-9}{2(y+3)}$$

Step 5.3.2.2:

Cancel the $(y+3)$.

$$\frac{y\cancel{(y+3)}}{\cancel{(y+3)}}=\frac{-9}{2(y+3)}$$

Step 5.3.3:

Simplify the right side.

$$y=-\frac{9}{2(y+3)}$$

Step 6:

Replace $y$ with $\frac{dy}{dx}$ to find the derivative.

$$\frac{dy}{dx}=-\frac{9}{2(y+3)}$$

Knowledge Notes:

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

2. Chain Rule: Used to differentiate composite functions. If $f(x)=h(g(x))$, then $f^{\prime }(x)=h^{\prime }(g(x))g^{\prime }(x)$.

3. Power Rule: The derivative of $x^n$ is $n{x}^{n-1}$.

4. Constant Multiple Rule: The derivative of a constant times a function is the constant times the derivative of the function.

5. Simplification: After differentiation, algebraic simplification may be necessary to isolate the variable of interest, in this case, $\frac{dy}{dx}$.

6. Implicit Differentiation: When differentiating an equation with respect to $x$ that involves $y$, and $y$ is a function of $x$, the derivative of $y$ with respect to $x$ is denoted as $\frac{dy}{dx}$.