Calculus Examples

$x^{2} - x p + p^{3} = 8$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d p} (x^{2} - x p + p^{3}) = \frac{d}{d p} (8)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

Differentiate using the chain rule, which states that $\frac{d}{d p} [f (g (p))]$ is $f' (g (p)) g' (p)$ where $f (p) = p^{2}$ and $g (p) = x$ .

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Step 2.2.1.1

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

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

Step 2.2.1.2

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

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

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Differentiate using the Product Rule which states that $\frac{d}{d p} [f (p) g (p)]$ is $f (p) \frac{d}{d p} [g (p)] + g (p) \frac{d}{d p} [f (p)]$ where $f (p) = x$ and $g (p) = p$ .

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $x$ by $1$ .

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

Step 2.4

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

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

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Remove unnecessary parentheses.

$2 x x' - x - p x' + 3 p^{2}$

Step 2.5.3

Reorder terms.

$3 p^{2} + 2 x x' - p x' - x$

Step 3

Since $8$ is constant with respect to $p$ , the derivative of $8$ with respect to $p$ is $0$ .

$0$

Step 4

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

$3 p^{2} + 2 x x' - p x' - x = 0$

Step 5

Solve for $x'$ .

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Step 5.1

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

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Step 5.1.1

Subtract $3 p^{2}$ from both sides of the equation.

$2 x x' - p x' - x = - 3 p^{2}$

Step 5.1.2

Add $x$ to both sides of the equation.

$2 x x' - p x' = - 3 p^{2} + x$

Step 5.2

Factor $x'$ out of $2 x x' - p x'$ .

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Step 5.2.1

Factor $x'$ out of $2 x x'$ .

$x' (2 x) - p x' = - 3 p^{2} + x$

Step 5.2.2

Factor $x'$ out of $- p x'$ .

$x' (2 x) + x' (- p) = - 3 p^{2} + x$

Step 5.2.3

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

$x' (2 x - p) = - 3 p^{2} + x$

Step 5.3

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

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Step 5.3.1

Divide each term in $x' (2 x - p) = - 3 p^{2} + x$ by $2 x - p$ .

$\frac{x' (2 x - p)}{2 x - p} = \frac{- 3 p^{2}}{2 x - p} + \frac{x}{2 x - p}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $2 x - p$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{x' (2 x - p)}{2 x - p} = \frac{- 3 p^{2}}{2 x - p} + \frac{x}{2 x - p}$

Step 5.3.2.1.2

Divide $x'$ by $1$ .

$x' = \frac{- 3 p^{2}}{2 x - p} + \frac{x}{2 x - p}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$x' = \frac{- 3 p^{2} + x}{2 x - p}$

Step 5.3.3.2

Factor $- 1$ out of $- 3 p^{2}$ .

$x' = \frac{- (3 p^{2}) + x}{2 x - p}$

Step 5.3.3.3

Factor $- 1$ out of $x$ .

$x' = \frac{- (3 p^{2}) - 1 (- x)}{2 x - p}$

Step 5.3.3.4

Factor $- 1$ out of $- (3 p^{2}) - 1 (- x)$ .

$x' = \frac{- (3 p^{2} - x)}{2 x - p}$

Step 5.3.3.5

Simplify the expression.

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Step 5.3.3.5.1

Rewrite $- (3 p^{2} - x)$ as $- 1 (3 p^{2} - x)$ .

$x' = \frac{- 1 (3 p^{2} - x)}{2 x - p}$

Step 5.3.3.5.2

Move the negative in front of the fraction.

$x' = - \frac{3 p^{2} - x}{2 x - p}$

Step 6

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

$\frac{d x}{d p} = - \frac{3 p^{2} - x}{2 x - p}$