Find dy/dx xy^4+x^5y=x+8y

$x y^{4} + x^{5} y = x + 8 y$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x y^{4} + x^{5} y) = \frac{d}{d x} (x + 8 y)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $x y^{4} + x^{5} y$ with respect to $x$ is $\frac{d}{d x} [x y^{4}] + \frac{d}{d x} [x^{5} y]$ .

$\frac{d}{d x} [x y^{4}] + \frac{d}{d x} [x^{5} y]$

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

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 $g (x) = y^{4}$ .

$x \frac{d}{d x} [y^{4}] + y^{4} \frac{d}{d x} [x] + \frac{d}{d x} [x^{5} y]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = y$ .

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

$x (\frac{d}{d u} [u^{4}] \frac{d}{d x} [y]) + y^{4} \frac{d}{d x} [x] + \frac{d}{d x} [x^{5} y]$

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

$x (4 u^{3} \frac{d}{d x} [y]) + y^{4} \frac{d}{d x} [x] + \frac{d}{d x} [x^{5} y]$

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

$x (4 y^{3} \frac{d}{d x} [y]) + y^{4} \frac{d}{d x} [x] + \frac{d}{d x} [x^{5} y]$

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

$x (4 y^{3} y') + y^{4} \frac{d}{d x} [x] + \frac{d}{d x} [x^{5} y]$

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

$x (4 y^{3} y') + y^{4} \cdot 1 + \frac{d}{d x} [x^{5} y]$

Move $4$ to the left of $x$ .

$4 \cdot x y^{3} y' + y^{4} \cdot 1 + \frac{d}{d x} [x^{5} y]$

Multiply $y^{4}$ by $1$ .

$4 x y^{3} y' + y^{4} + \frac{d}{d x} [x^{5} y]$

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

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 $f (x) = x^{5}$ and $g (x) = y$ .

$4 x y^{3} y' + y^{4} + x^{5} \frac{d}{d x} [y] + y \frac{d}{d x} [x^{5}]$

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

$4 x y^{3} y' + y^{4} + x^{5} y' + y \frac{d}{d x} [x^{5}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$4 x y^{3} y' + y^{4} + x^{5} y' + y (5 x^{4})$

Move $5$ to the left of $y$ .

$4 x y^{3} y' + y^{4} + x^{5} y' + 5 y x^{4}$

Reorder terms.

$y^{4} + 4 y^{3} x y' + x^{5} y' + 5 x^{4} y$

Step 3

Differentiate the right side of the equation.

Differentiate.

By the Sum Rule, the derivative of $x + 8 y$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [8 y]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [8 y]$

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

$1 + \frac{d}{d x} [8 y]$

Evaluate $\frac{d}{d x} [8 y]$ .

Since $8$ is constant with respect to $x$ , the derivative of $8 y$ with respect to $x$ is $8 \frac{d}{d x} [y]$ .

$1 + 8 \frac{d}{d x} [y]$

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

$1 + 8 y'$

Reorder terms.

$8 y' + 1$

Step 4

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

$y^{4} + 4 y^{3} x y' + x^{5} y' + 5 x^{4} y = 8 y' + 1$

Step 5

Solve for $y'$ .

Subtract $8 y'$ from both sides of the equation.

$y^{4} + 4 y^{3} x y' + x^{5} y' + 5 x^{4} y - 8 y' = 1$

Move all terms not containing $y'$ to the right side of the equation.

Subtract $y^{4}$ from both sides of the equation.

$4 y^{3} x y' + x^{5} y' + 5 x^{4} y - 8 y' = 1 - y^{4}$

Subtract $5 x^{4} y$ from both sides of the equation.

$4 y^{3} x y' + x^{5} y' - 8 y' = 1 - y^{4} - 5 x^{4} y$

Factor $y'$ out of $4 y^{3} x y' + x^{5} y' - 8 y'$ .

Factor $y'$ out of $4 y^{3} x y'$ .

$y' (4 y^{3} x) + x^{5} y' - 8 y' = 1 - y^{4} - 5 x^{4} y$

Factor $y'$ out of $x^{5} y'$ .

$y' (4 y^{3} x) + y' (x^{5}) - 8 y' = 1 - y^{4} - 5 x^{4} y$

Factor $y'$ out of $- 8 y'$ .

$y' (4 y^{3} x) + y' (x^{5}) + y' \cdot - 8 = 1 - y^{4} - 5 x^{4} y$

Factor $y'$ out of $y' (4 y^{3} x) + y' (x^{5})$ .

$y' (4 y^{3} x + x^{5}) + y' \cdot - 8 = 1 - y^{4} - 5 x^{4} y$

Factor $y'$ out of $y' (4 y^{3} x + x^{5}) + y' \cdot - 8$ .

$y' (4 y^{3} x + x^{5} - 8) = 1 - y^{4} - 5 x^{4} y$

Divide each term in $y' (4 y^{3} x + x^{5} - 8) = 1 - y^{4} - 5 x^{4} y$ by $4 y^{3} x + x^{5} - 8$ and simplify.

Divide each term in $y' (4 y^{3} x + x^{5} - 8) = 1 - y^{4} - 5 x^{4} y$ by $4 y^{3} x + x^{5} - 8$ .

$\frac{y' (4 y^{3} x + x^{5} - 8)}{4 y^{3} x + x^{5} - 8} = \frac{1}{4 y^{3} x + x^{5} - 8} + \frac{- y^{4}}{4 y^{3} x + x^{5} - 8} + \frac{- 5 x^{4} y}{4 y^{3} x + x^{5} - 8}$

Simplify the left side.

Cancel the common factor of $4 y^{3} x + x^{5} - 8$ .

Cancel the common factor.

$\frac{y' (4 y^{3} x + x^{5} - 8)}{4 y^{3} x + x^{5} - 8} = \frac{1}{4 y^{3} x + x^{5} - 8} + \frac{- y^{4}}{4 y^{3} x + x^{5} - 8} + \frac{- 5 x^{4} y}{4 y^{3} x + x^{5} - 8}$

Divide $y'$ by $1$ .

$y' = \frac{1}{4 y^{3} x + x^{5} - 8} + \frac{- y^{4}}{4 y^{3} x + x^{5} - 8} + \frac{- 5 x^{4} y}{4 y^{3} x + x^{5} - 8}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y' = \frac{1}{4 y^{3} x + x^{5} - 8} - \frac{y^{4}}{4 y^{3} x + x^{5} - 8} + \frac{- 5 x^{4} y}{4 y^{3} x + x^{5} - 8}$

Move the negative in front of the fraction.

$y' = \frac{1}{4 y^{3} x + x^{5} - 8} - \frac{y^{4}}{4 y^{3} x + x^{5} - 8} - \frac{5 x^{4} y}{4 y^{3} x + x^{5} - 8}$

Combine into one fraction.

Combine the numerators over the common denominator.

$y' = \frac{1 - y^{4}}{4 y^{3} x + x^{5} - 8} - \frac{5 x^{4} y}{4 y^{3} x + x^{5} - 8}$

Combine the numerators over the common denominator.

$y' = \frac{1 - y^{4} - 5 x^{4} y}{4 y^{3} x + x^{5} - 8}$

Step 6

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

$\frac{d y}{d x} = \frac{1 - y^{4} - 5 x^{4} y}{4 y^{3} x + x^{5} - 8}$

Find dy/dx xy^4+x^5y=x+8y

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