Find dy/dx y^2+5y+6x=0

$y^{2} + 5 y + 6 x = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{2} + 5 y + 6 x) = \frac{d}{d x} (0)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $y^{2} + 5 y + 6 x$ with respect to $x$ is $\frac{d}{d x} [y^{2}] + \frac{d}{d x} [5 y] + \frac{d}{d x} [6 x]$ .

$\frac{d}{d x} [y^{2}] + \frac{d}{d x} [5 y] + \frac{d}{d x} [6 x]$

Evaluate $\frac{d}{d x} [y^{2}]$ .

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = y$ .

To apply the Chain Rule, set $u$ as $y$ .

$\frac{d}{d u} [u^{2}] \frac{d}{d x} [y] + \frac{d}{d x} [5 y] + \frac{d}{d x} [6 x]$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

$2 u \frac{d}{d x} [y] + \frac{d}{d x} [5 y] + \frac{d}{d x} [6 x]$

Replace all occurrences of $u$ with $y$ .

$2 y \frac{d}{d x} [y] + \frac{d}{d x} [5 y] + \frac{d}{d x} [6 x]$

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$2 y y' + \frac{d}{d x} [5 y] + \frac{d}{d x} [6 x]$

Evaluate $\frac{d}{d x} [5 y]$ .

Since $5$ is constant with respect to $x$ , the derivative of $5 y$ with respect to $x$ is $5 \frac{d}{d x} [y]$ .

$2 y y' + 5 \frac{d}{d x} [y] + \frac{d}{d x} [6 x]$

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$2 y y' + 5 y' + \frac{d}{d x} [6 x]$

Evaluate $\frac{d}{d x} [6 x]$ .

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$2 y y' + 5 y' + 6 \frac{d}{d x} [x]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 y y' + 5 y' + 6 \cdot 1$

Multiply $6$ by $1$ .

$2 y y' + 5 y' + 6$

Step 3

Since $0$ is constant with respect to $x$ , the derivative of $0$ with respect to $x$ is $0$ .

$0$

Step 4

Reform the equation by setting the left side equal to the right side.

$2 y y' + 5 y' + 6 = 0$

Step 5

Solve for $y'$ .

Subtract $6$ from both sides of the equation.

$2 y y' + 5 y' = - 6$

Factor $y'$ out of $2 y y' + 5 y'$ .

Factor $y'$ out of $2 y y'$ .

$y' (2 y) + 5 y' = - 6$

Factor $y'$ out of $5 y'$ .

$y' (2 y) + y' \cdot 5 = - 6$

Factor $y'$ out of $y' (2 y) + y' \cdot 5$ .

$y' (2 y + 5) = - 6$

Divide each term in $y' (2 y + 5) = - 6$ by $2 y + 5$ and simplify.

Divide each term in $y' (2 y + 5) = - 6$ by $2 y + 5$ .

$\frac{y' (2 y + 5)}{2 y + 5} = \frac{- 6}{2 y + 5}$

Simplify the left side.

Cancel the common factor of $2 y + 5$ .

Cancel the common factor.

$\frac{y' (2 y + 5)}{2 y + 5} = \frac{- 6}{2 y + 5}$

Divide $y'$ by $1$ .

$y' = \frac{- 6}{2 y + 5}$

Simplify the right side.

Move the negative in front of the fraction.

$y' = - \frac{6}{2 y + 5}$

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - \frac{6}{2 y + 5}$

Find dy/dx y^2+5y+6x=0

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