Calculus Examples

$4 y^{2} + x y - 2 x^{2} = 3$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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 $g (x) = y$ .

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $2$ by $4$ .

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

Step 2.3

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

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Step 2.3.1

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 $g (x) = y$ .

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Multiply $y$ by $1$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Multiply $2$ by $- 2$ .

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

Step 2.5

Reorder terms.

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

Step 3

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$0$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

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

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Step 5.1.1

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

$8 y y' + x y' + y = 4 x$

Step 5.1.2

Subtract $y$ from both sides of the equation.

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

Step 5.2

Factor $y'$ out of $8 y y' + x y'$ .

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

Factor $y'$ out of $y' (8 y) + y' (x)$ .

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $8 y + x$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

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

Step 6

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

$\frac{d y}{d x} = \frac{4 x - y}{8 y + x}$