Calculus Examples

${(2 x - y)}^{2} - 3 y = 2$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

Tap for more steps...

Step 2.1

Rewrite ${(2 x - y)}^{2}$ as $(2 x - y) (2 x - y)$ .

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

Step 2.2

Expand $(2 x - y) (2 x - y)$ using the FOIL Method.

Tap for more steps...

Step 2.2.1

Apply the distributive property.

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

Step 2.2.2

Apply the distributive property.

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

Step 2.2.3

Apply the distributive property.

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

Step 2.3

Simplify and combine like terms.

Tap for more steps...

Step 2.3.1

Simplify each term.

Tap for more steps...

Step 2.3.1.1

Rewrite using the commutative property of multiplication.

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

Step 2.3.1.2

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 2.3.1.2.1

Move $x$ .

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

Step 2.3.1.2.2

Multiply $x$ by $x$ .

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

Step 2.3.1.3

Multiply $2$ by $2$ .

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

Step 2.3.1.4

Rewrite using the commutative property of multiplication.

$\frac{d}{d x} [4 x^{2} + 2 \cdot - 1 x y - y (2 x) - y (- y) - 3 y]$

Step 2.3.1.5

Multiply $2$ by $- 1$ .

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

Step 2.3.1.6

Rewrite using the commutative property of multiplication.

$\frac{d}{d x} [4 x^{2} - 2 x y - 1 \cdot 2 y x - y (- y) - 3 y]$

Step 2.3.1.7

Multiply $- 1$ by $2$ .

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

Step 2.3.1.8

Rewrite using the commutative property of multiplication.

$\frac{d}{d x} [4 x^{2} - 2 x y - 2 y x - 1 \cdot - 1 y \cdot y - 3 y]$

Step 2.3.1.9

Multiply $y$ by $y$ by adding the exponents.

Tap for more steps...

Step 2.3.1.9.1

Move $y$ .

$\frac{d}{d x} [4 x^{2} - 2 x y - 2 y x - 1 \cdot - 1 (y \cdot y) - 3 y]$

Step 2.3.1.9.2

Multiply $y$ by $y$ .

$\frac{d}{d x} [4 x^{2} - 2 x y - 2 y x - 1 \cdot - 1 y^{2} - 3 y]$

Step 2.3.1.10

Multiply $- 1$ by $- 1$ .

$\frac{d}{d x} [4 x^{2} - 2 x y - 2 y x + 1 y^{2} - 3 y]$

Step 2.3.1.11

Multiply $y^{2}$ by $1$ .

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

Step 2.3.2

Subtract $2 y x$ from $- 2 x y$ .

Tap for more steps...

Step 2.3.2.1

Move $y$ .

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

Step 2.3.2.2

Subtract $2 x y$ from $- 2 x y$ .

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

Step 2.4

By the Sum Rule, the derivative of $4 x^{2} - 4 x y + y^{2} - 3 y$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 4 x y] + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 3 y]$ .

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

Step 2.5

Evaluate $\frac{d}{d x} [4 x^{2}]$ .

Tap for more steps...

Step 2.5.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

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

Step 2.5.2

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

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

Step 2.5.3

Multiply $2$ by $4$ .

$8 x + \frac{d}{d x} [- 4 x y] + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 3 y]$

Step 2.6

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

Tap for more steps...

Step 2.6.1

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

$8 x - 4 \frac{d}{d x} [x y] + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 3 y]$

Step 2.6.2

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 $g (x) = y$ .

$8 x - 4 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x]) + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 3 y]$

Step 2.6.3

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

$8 x - 4 (x y' + y \frac{d}{d x} [x]) + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 3 y]$

Step 2.6.4

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

$8 x - 4 (x y' + y \cdot 1) + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 3 y]$

Step 2.6.5

Multiply $y$ by $1$ .

$8 x - 4 (x y' + y) + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 3 y]$

Step 2.7

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

Tap for more steps...

Step 2.7.1

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 $g (x) = y$ .

Tap for more steps...

Step 2.7.1.1

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

$8 x - 4 (x y' + y) + \frac{d}{d u} [u^{2}] \frac{d}{d x} [y] + \frac{d}{d x} [- 3 y]$

Step 2.7.1.2

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

$8 x - 4 (x y' + y) + 2 u \frac{d}{d x} [y] + \frac{d}{d x} [- 3 y]$

Step 2.7.1.3

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

$8 x - 4 (x y' + y) + 2 y \frac{d}{d x} [y] + \frac{d}{d x} [- 3 y]$

Step 2.7.2

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

$8 x - 4 (x y' + y) + 2 y y' + \frac{d}{d x} [- 3 y]$

Step 2.8

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

Tap for more steps...

Step 2.8.1

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

$8 x - 4 (x y' + y) + 2 y y' - 3 \frac{d}{d x} [y]$

Step 2.8.2

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

$8 x - 4 (x y' + y) + 2 y y' - 3 y'$

Step 2.9

Simplify.

Tap for more steps...

Step 2.9.1

Apply the distributive property.

$8 x - 4 (x y') - 4 y + 2 y y' - 3 y'$

Step 2.9.2

Remove unnecessary parentheses.

$8 x - 4 x y' - 4 y + 2 y y' - 3 y'$

Step 2.9.3

Reorder terms.

$- 4 x y' + 2 y y' + 8 x - 4 y - 3 y'$

Step 3

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$0$

Step 4

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

$- 4 x y' + 2 y y' + 8 x - 4 y - 3 y' = 0$

Step 5

Solve for $y'$ .

Tap for more steps...

Step 5.1

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

Tap for more steps...

Step 5.1.1

Subtract $8 x$ from both sides of the equation.

$- 4 x y' + 2 y y' - 4 y - 3 y' = - 8 x$

Step 5.1.2

Add $4 y$ to both sides of the equation.

$- 4 x y' + 2 y y' - 3 y' = - 8 x + 4 y$

Step 5.2

Factor $y'$ out of $- 4 x y' + 2 y y' - 3 y'$ .

Tap for more steps...

Step 5.2.1

Factor $y'$ out of $- 4 x y'$ .

$y' (- 4 x) + 2 y y' - 3 y' = - 8 x + 4 y$

Step 5.2.2

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

$y' (- 4 x) + y' (2 y) - 3 y' = - 8 x + 4 y$

Step 5.2.3

Factor $y'$ out of $- 3 y'$ .

$y' (- 4 x) + y' (2 y) + y' \cdot - 3 = - 8 x + 4 y$

Step 5.2.4

Factor $y'$ out of $y' (- 4 x) + y' (2 y)$ .

$y' (- 4 x + 2 y) + y' \cdot - 3 = - 8 x + 4 y$

Step 5.2.5

Factor $y'$ out of $y' (- 4 x + 2 y) + y' \cdot - 3$ .

$y' (- 4 x + 2 y - 3) = - 8 x + 4 y$

Step 5.3

Divide each term in $y' (- 4 x + 2 y - 3) = - 8 x + 4 y$ by $- 4 x + 2 y - 3$ and simplify.

Tap for more steps...

Step 5.3.1

Divide each term in $y' (- 4 x + 2 y - 3) = - 8 x + 4 y$ by $- 4 x + 2 y - 3$ .

$\frac{y' (- 4 x + 2 y - 3)}{- 4 x + 2 y - 3} = \frac{- 8 x}{- 4 x + 2 y - 3} + \frac{4 y}{- 4 x + 2 y - 3}$

Step 5.3.2

Simplify the left side.

Tap for more steps...

Step 5.3.2.1

Cancel the common factor of $- 4 x + 2 y - 3$ .

Tap for more steps...

Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (- 4 x + 2 y - 3)}{- 4 x + 2 y - 3} = \frac{- 8 x}{- 4 x + 2 y - 3} + \frac{4 y}{- 4 x + 2 y - 3}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 8 x}{- 4 x + 2 y - 3} + \frac{4 y}{- 4 x + 2 y - 3}$

Step 5.3.3

Simplify the right side.

Tap for more steps...

Step 5.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{- 8 x + 4 y}{- 4 x + 2 y - 3}$

Step 5.3.3.2

Factor $4$ out of $- 8 x + 4 y$ .

Tap for more steps...

Step 5.3.3.2.1

Factor $4$ out of $- 8 x$ .

$y' = \frac{4 (- 2 x) + 4 y}{- 4 x + 2 y - 3}$

Step 5.3.3.2.2

Factor $4$ out of $4 y$ .

$y' = \frac{4 (- 2 x) + 4 (y)}{- 4 x + 2 y - 3}$

Step 5.3.3.2.3

Factor $4$ out of $4 (- 2 x) + 4 (y)$ .

$y' = \frac{4 (- 2 x + y)}{- 4 x + 2 y - 3}$

Step 5.3.3.3

Factor $- 1$ out of $- 2 x$ .

$y' = \frac{4 (- (2 x) + y)}{- 4 x + 2 y - 3}$

Step 5.3.3.4

Factor $- 1$ out of $y$ .

$y' = \frac{4 (- (2 x) - 1 (- y))}{- 4 x + 2 y - 3}$

Step 5.3.3.5

Factor $- 1$ out of $- (2 x) - 1 (- y)$ .

$y' = \frac{4 (- (2 x - y))}{- 4 x + 2 y - 3}$

Step 5.3.3.6

Rewrite $- (2 x - y)$ as $- 1 (2 x - y)$ .

$y' = \frac{4 (- 1 (2 x - y))}{- 4 x + 2 y - 3}$

Step 5.3.3.7

Factor $- 1$ out of $- 4 x$ .

$y' = \frac{4 (- 1 (2 x - y))}{- (4 x) + 2 y - 3}$

Step 5.3.3.8

Factor $- 1$ out of $2 y$ .

$y' = \frac{4 (- 1 (2 x - y))}{- (4 x) - (- 2 y) - 3}$

Step 5.3.3.9

Factor $- 1$ out of $- (4 x) - (- 2 y)$ .

$y' = \frac{4 (- 1 (2 x - y))}{- (4 x - 2 y) - 3}$

Step 5.3.3.10

Rewrite $- 3$ as $- 1 (3)$ .

$y' = \frac{4 (- 1 (2 x - y))}{- (4 x - 2 y) - 1 (3)}$

Step 5.3.3.11

Factor $- 1$ out of $- (4 x - 2 y) - 1 (3)$ .

$y' = \frac{4 (- 1 (2 x - y))}{- (4 x - 2 y + 3)}$

Step 5.3.3.12

Rewrite $- (4 x - 2 y + 3)$ as $- 1 (4 x - 2 y + 3)$ .

$y' = \frac{4 (- 1 (2 x - y))}{- 1 (4 x - 2 y + 3)}$

Step 5.3.3.13

Cancel the common factor.

$y' = \frac{4 (- 1 (2 x - y))}{- 1 (4 x - 2 y + 3)}$

Step 5.3.3.14

Rewrite the expression.

$y' = \frac{4 (2 x - y)}{4 x - 2 y + 3}$

Step 6

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

$\frac{d y}{d x} = \frac{4 (2 x - y)}{4 x - 2 y + 3}$