Find dy/dx 4y^3+xy-y=32x^4

$4 y^{3} + x y - y = 32 x^{4}$

Differentiate both sides of the equation.

$\frac{d}{d x} (4 y^{3} + x y - y) = \frac{d}{d x} (32 x^{4})$

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $4 y^{3} + x y - y$ with respect to $x$ is $\frac{d}{d x} [4 y^{3}] + \frac{d}{d x} [x y] + \frac{d}{d x} [- y]$ .

$\frac{d}{d x} [4 y^{3}] + \frac{d}{d x} [x y] + \frac{d}{d x} [- y]$

Evaluate $\frac{d}{d x} [4 y^{3}]$ .

Since $4$ is constant with respect to $x$ , the derivative of $4 y^{3}$ with respect to $x$ is $4 \frac{d}{d x} [y^{3}]$ .

$4 \frac{d}{d x} [y^{3}] + \frac{d}{d x} [x y] + \frac{d}{d x} [- y]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = y$ .

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

$4 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y]) + \frac{d}{d x} [x y] + \frac{d}{d x} [- y]$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$4 (3 u^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [x y] + \frac{d}{d x} [- y]$

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

$4 (3 y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [x y] + \frac{d}{d x} [- y]$

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

$4 (3 y^{2} y') + \frac{d}{d x} [x y] + \frac{d}{d x} [- y]$

Multiply $3$ by $4$ .

$12 y^{2} y' + \frac{d}{d x} [x y] + \frac{d}{d x} [- y]$

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

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 $f (x) = x$ and $g (x) = y$ .

$12 y^{2} y' + x \frac{d}{d x} [y] + y \frac{d}{d x} [x] + \frac{d}{d x} [- y]$

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

$12 y^{2} y' + x y' + y \frac{d}{d x} [x] + \frac{d}{d x} [- y]$

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

$12 y^{2} y' + x y' + y \cdot 1 + \frac{d}{d x} [- y]$

Multiply $y$ by $1$ .

$12 y^{2} y' + x y' + y + \frac{d}{d x} [- y]$

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- y$ with respect to $x$ is $- \frac{d}{d x} [y]$ .

$12 y^{2} y' + x y' + y - \frac{d}{d x} [y]$

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

$12 y^{2} y' + x y' + y - y'$

Differentiate the right side of the equation.

Since $32$ is constant with respect to $x$ , the derivative of $32 x^{4}$ with respect to $x$ is $32 \frac{d}{d x} [x^{4}]$ .

$32 \frac{d}{d x} [x^{4}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$32 (4 x^{3})$

Multiply $4$ by $32$ .

$128 x^{3}$

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

$12 y^{2} y' + x y' + y - y' = 128 x^{3}$

Solve for $y'$ .

Subtract $y$ from both sides of the equation.

$12 y^{2} y' + x y' - y' = 128 x^{3} - y$

Factor $y'$ out of $12 y^{2} y' + x y' - y'$ .

Factor $y'$ out of $12 y^{2} y'$ .

$y' (12 y^{2}) + x (y') - y' = 128 x^{3} - y$

Factor $y'$ out of $x y'$ .

$y' (12 y^{2}) + y' (x) - y' = 128 x^{3} - y$

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

$y' (12 y^{2}) + y' (x) + y' \cdot - 1 = 128 x^{3} - y$

Factor $y'$ out of $y' (12 y^{2}) + y' (x)$ .

$y' (12 y^{2} + x) + y' \cdot - 1 = 128 x^{3} - y$

Factor $y'$ out of $y' (12 y^{2} + x) + y' \cdot - 1$ .

$y' (12 y^{2} + x - 1) = 128 x^{3} - y$

Divide each term in $y' (12 y^{2} + x - 1) = 128 x^{3} - y$ by $12 y^{2} + x - 1$ and simplify.

Divide each term in $y' (12 y^{2} + x - 1) = 128 x^{3} - y$ by $12 y^{2} + x - 1$ .

$\frac{y' (12 y^{2} + x - 1)}{12 y^{2} + x - 1} = \frac{128 x^{3}}{12 y^{2} + x - 1} + \frac{- y}{12 y^{2} + x - 1}$

Simplify the left side.

Cancel the common factor of $12 y^{2} + x - 1$ .

Cancel the common factor.

$\frac{y' (12 y^{2} + x - 1)}{12 y^{2} + x - 1} = \frac{128 x^{3}}{12 y^{2} + x - 1} + \frac{- y}{12 y^{2} + x - 1}$

Divide $y'$ by $1$ .

$y' = \frac{128 x^{3}}{12 y^{2} + x - 1} + \frac{- y}{12 y^{2} + x - 1}$

Simplify the right side.

Combine the numerators over the common denominator.

$y' = \frac{128 x^{3} - y}{12 y^{2} + x - 1}$

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

$\frac{d y}{d x} = \frac{128 x^{3} - y}{12 y^{2} + x - 1}$

Find dy/dx 4y^3+xy-y=32x^4

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