Find dy/dx 3x^2y-picos(y)=4pi

$3 x^{2} y - π \cos (y) = 4 π$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (3 x^{2} y - π \cos (y)) = \frac{d}{d x} (4 π)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $3 x^{2} y - π \cos (y)$ with respect to $x$ is $\frac{d}{d x} [3 x^{2} y] + \frac{d}{d x} [- π \cos (y)]$ .

$\frac{d}{d x} [3 x^{2} y] + \frac{d}{d x} [- π \cos (y)]$

Evaluate $\frac{d}{d x} [3 x^{2} y]$ .

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2} y$ with respect to $x$ is $3 \frac{d}{d x} [x^{2} y]$ .

$3 \frac{d}{d x} [x^{2} y] + \frac{d}{d x} [- π \cos (y)]$

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{2}$ and $g (x) = y$ .

$3 (x^{2} \frac{d}{d x} [y] + y \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [- π \cos (y)]$

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

$3 (x^{2} y' + y \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [- π \cos (y)]$

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

$3 (x^{2} y' + y (2 x)) + \frac{d}{d x} [- π \cos (y)]$

Move $2$ to the left of $y$ .

$3 (x^{2} y' + 2 y x) + \frac{d}{d x} [- π \cos (y)]$

Evaluate $\frac{d}{d x} [- π \cos (y)]$ .

Since $- π$ is constant with respect to $x$ , the derivative of $- π \cos (y)$ with respect to $x$ is $- π \frac{d}{d x} [\cos (y)]$ .

$3 (x^{2} y' + 2 y x) - π \frac{d}{d x} [\cos (y)]$

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

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

$3 (x^{2} y' + 2 y x) - π (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [y])$

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

$3 (x^{2} y' + 2 y x) - π (- \sin (u) \frac{d}{d x} [y])$

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

$3 (x^{2} y' + 2 y x) - π (- \sin (y) \frac{d}{d x} [y])$

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

$3 (x^{2} y' + 2 y x) - π (- \sin (y) y')$

Multiply $- 1$ by $- 1$ .

$3 (x^{2} y' + 2 y x) + 1 π (\sin (y) y')$

Multiply $π$ by $1$ .

$3 (x^{2} y' + 2 y x) + π \sin (y) y'$

Simplify.

Apply the distributive property.

$3 (x^{2} y') + 3 (2 y x) + π \sin (y) y'$

Multiply $2$ by $3$ .

$3 x^{2} y' + 6 y x + π \sin (y) y'$

Reorder terms.

$3 x^{2} y' + 6 x y + π \sin (y) y'$

Step 3

Since $4 π$ is constant with respect to $x$ , the derivative of $4 π$ with respect to $x$ is $0$ .

$0$

Step 4

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

$3 x^{2} y' + 6 x y + π \sin (y) y' = 0$

Step 5

Solve for $y'$ .

Simplify the left side.

Reorder factors in $3 x^{2} y' + 6 x y + π \sin (y) y'$ .

$3 x^{2} y' + 6 x y + π y' \sin (y) = 0$

Subtract $6 x y$ from both sides of the equation.

$3 x^{2} y' + π y' \sin (y) = - 6 x y$

Factor $y'$ out of $3 x^{2} y' + π y' \sin (y)$ .

Factor $y'$ out of $3 x^{2} y'$ .

$y' (3 x^{2}) + π y' \sin (y) = - 6 x y$

Factor $y'$ out of $π y' \sin (y)$ .

$y' (3 x^{2}) + y' (π \sin (y)) = - 6 x y$

Factor $y'$ out of $y' (3 x^{2}) + y' (π \sin (y))$ .

$y' (3 x^{2} + π \sin (y)) = - 6 x y$

Divide each term in $y' (3 x^{2} + π \sin (y)) = - 6 x y$ by $3 x^{2} + π \sin (y)$ and simplify.

Divide each term in $y' (3 x^{2} + π \sin (y)) = - 6 x y$ by $3 x^{2} + π \sin (y)$ .

$\frac{y' (3 x^{2} + π \sin (y))}{3 x^{2} + π \sin (y)} = \frac{- 6 x y}{3 x^{2} + π \sin (y)}$

Simplify the left side.

Cancel the common factor of $3 x^{2} + π \sin (y)$ .

Cancel the common factor.

$\frac{y' (3 x^{2} + π \sin (y))}{3 x^{2} + π \sin (y)} = \frac{- 6 x y}{3 x^{2} + π \sin (y)}$

Divide $y'$ by $1$ .

$y' = \frac{- 6 x y}{3 x^{2} + π \sin (y)}$

Simplify the right side.

Move the negative in front of the fraction.

$y' = - \frac{6 x y}{3 x^{2} + π \sin (y)}$

Step 6

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

$\frac{d y}{d x} = - \frac{6 x y}{3 x^{2} + π \sin (y)}$

Find dy/dx 3x^2y-picos(y)=4pi

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