Calculus Examples

$3 x^{2} y^{3} - x + y = 25$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (3 x^{2} y^{3} - x + y) = \frac{d}{d x} (25)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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 $f (x) = x^{2}$ and $g (x) = y^{3}$ .

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

Step 2.2.3

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 $g (x) = y$ .

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Step 2.2.3.1

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

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

Step 2.2.3.2

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

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

Step 2.2.3.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

Move $3$ to the left of $x^{2}$ .

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

Step 2.2.7

Move $2$ to the left of $y^{3}$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 1$ by $1$ .

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

Step 2.4

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

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

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Combine terms.

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Step 2.5.2.1

Multiply $3$ by $3$ .

$9 (x^{2} y^{2} y') + 3 (2 y^{3} x) - 1 + y'$

Step 2.5.2.2

Multiply $2$ by $3$ .

$9 x^{2} y^{2} y' + 6 y^{3} x - 1 + y'$

Step 3

Since $25$ is constant with respect to $x$ , the derivative of $25$ with respect to $x$ is $0$ .

$0$

Step 4

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

$9 x^{2} y^{2} y' + 6 y^{3} x - 1 + y' = 0$

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

Subtract $6 y^{3} x$ from both sides of the equation.

$9 x^{2} y^{2} y' - 1 + y' = - 6 y^{3} x$

Step 5.1.2

Add $1$ to both sides of the equation.

$9 x^{2} y^{2} y' + y' = - 6 y^{3} x + 1$

Step 5.2

Factor $y'$ out of $9 x^{2} y^{2} y' + y'$ .

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Step 5.2.1

Factor $y'$ out of $9 x^{2} y^{2} y'$ .

$y' (9 x^{2} y^{2}) + y' = - 6 y^{3} x + 1$

Step 5.2.2

Raise $y'$ to the power of $1$ .

$y' (9 x^{2} y^{2}) + y' = - 6 y^{3} x + 1$

Step 5.2.3

Factor $y'$ out of ${y'}^{1}$ .

$y' (9 x^{2} y^{2}) + y' \cdot 1 = - 6 y^{3} x + 1$

Step 5.2.4

Factor $y'$ out of $y' (9 x^{2} y^{2}) + y' \cdot 1$ .

$y' (9 x^{2} y^{2} + 1) = - 6 y^{3} x + 1$

Step 5.3

Divide each term in $y' (9 x^{2} y^{2} + 1) = - 6 y^{3} x + 1$ by $9 x^{2} y^{2} + 1$ and simplify.

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Step 5.3.1

Divide each term in $y' (9 x^{2} y^{2} + 1) = - 6 y^{3} x + 1$ by $9 x^{2} y^{2} + 1$ .

$\frac{y' (9 x^{2} y^{2} + 1)}{9 x^{2} y^{2} + 1} = \frac{- 6 y^{3} x}{9 x^{2} y^{2} + 1} + \frac{1}{9 x^{2} y^{2} + 1}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $9 x^{2} y^{2} + 1$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (9 x^{2} y^{2} + 1)}{9 x^{2} y^{2} + 1} = \frac{- 6 y^{3} x}{9 x^{2} y^{2} + 1} + \frac{1}{9 x^{2} y^{2} + 1}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 6 y^{3} x}{9 x^{2} y^{2} + 1} + \frac{1}{9 x^{2} y^{2} + 1}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{- 6 y^{3} x + 1}{9 x^{2} y^{2} + 1}$

Step 5.3.3.2

Factor $- 1$ out of $- 6 y^{3} x$ .

$y' = \frac{- (6 y^{3} x) + 1}{9 x^{2} y^{2} + 1}$

Step 5.3.3.3

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

$y' = \frac{- (6 y^{3} x) - 1 (- 1)}{9 x^{2} y^{2} + 1}$

Step 5.3.3.4

Factor $- 1$ out of $- (6 y^{3} x) - 1 (- 1)$ .

$y' = \frac{- (6 y^{3} x - 1)}{9 x^{2} y^{2} + 1}$

Step 5.3.3.5

Simplify the expression.

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Step 5.3.3.5.1

Rewrite $- (6 y^{3} x - 1)$ as $- 1 (6 y^{3} x - 1)$ .

$y' = \frac{- 1 (6 y^{3} x - 1)}{9 x^{2} y^{2} + 1}$

Step 5.3.3.5.2

Move the negative in front of the fraction.

$y' = - \frac{6 y^{3} x - 1}{9 x^{2} y^{2} + 1}$

Step 6

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

$\frac{d y}{d x} = - \frac{6 y^{3} x - 1}{9 x^{2} y^{2} + 1}$