Find dy/dx sin(y)+6x=y^2

$\sin (y) + 6 x = y^{2}$

Differentiate both sides of the equation.

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

Differentiate the left side of the equation.

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

$\frac{d}{d x} [\sin (y)] + \frac{d}{d x} [6 x]$

Evaluate $\frac{d}{d x} [\sin (y)]$ .

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

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

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [y] + \frac{d}{d x} [6 x]$

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$\cos (u) \frac{d}{d x} [y] + \frac{d}{d x} [6 x]$

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

$\cos (y) \frac{d}{d x} [y] + \frac{d}{d x} [6 x]$

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

$\cos (y) y' + \frac{d}{d x} [6 x]$

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

$\cos (y) y' + 6 \frac{d}{d x} [x]$

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

$\cos (y) y' + 6 \cdot 1$

Multiply $6$ by $1$ .

$\cos (y) y' + 6$

Differentiate the right side of the equation.

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 $y$ .

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

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

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

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

$2 y \frac{d}{d x} [y]$

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

$2 y y'$

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

$\cos (y) y' + 6 = 2 y y'$

Solve for $y'$ .

Simplify the left side.

Reorder factors in $\cos (y) y' + 6$ .

$y' \cos (y) + 6 = 2 y y'$

Subtract $2 y y'$ from both sides of the equation.

$y' \cos (y) + 6 - 2 y y' = 0$

Subtract $6$ from both sides of the equation.

$y' \cos (y) - 2 y y' = - 6$

Factor $y'$ out of $y' \cos (y) - 2 y y'$ .

Factor $y'$ out of $y' \cos (y)$ .

$y' (\cos (y)) - 2 y y' = - 6$

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

$y' (\cos (y)) + y' (- 2 y) = - 6$

Factor $y'$ out of $y' (\cos (y)) + y' (- 2 y)$ .

$y' (\cos (y) - 2 y) = - 6$

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

Divide each term in $y' (\cos (y) - 2 y) = - 6$ by $\cos (y) - 2 y$ .

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

Simplify the left side.

Cancel the common factor of $\cos (y) - 2 y$ .

Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

$\frac{d y}{d x} = - \frac{6}{\cos (y) - 2 y}$

Find dy/dx sin(y)+6x=y^2

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