Find dx/dp x^2-3xp+p^2=5

$x^{2} - 3 x p + p^{2} = 5$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d p} (x^{2} - 3 x p + p^{2}) = \frac{d}{d p} (5)$

Step 2

Differentiate the left side of the equation.

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

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

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

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 $g (p) = x$ .

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

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

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

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

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

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

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

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

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

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

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

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 $g (p) = p$ .

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

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

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

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

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

Multiply $x$ by $1$ .

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

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

$2 x x' - 3 (x + p x') + 2 p$

Simplify.

Apply the distributive property.

$2 x x' - 3 x - 3 (p x') + 2 p$

Remove unnecessary parentheses.

$2 x x' - 3 x - 3 p x' + 2 p$

Reorder terms.

$2 x x' - 3 p x' + 2 p - 3 x$

Step 3

Since $5$ is constant with respect to $p$ , the derivative of $5$ with respect to $p$ is $0$ .

$0$

Step 4

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

$2 x x' - 3 p x' + 2 p - 3 x = 0$

Step 5

Solve for $x'$ .

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

Subtract $2 p$ from both sides of the equation.

$2 x x' - 3 p x' - 3 x = - 2 p$

Add $3 x$ to both sides of the equation.

$2 x x' - 3 p x' = - 2 p + 3 x$

Factor $x'$ out of $2 x x' - 3 p x'$ .

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

$x' (2 x) - 3 p x' = - 2 p + 3 x$

Factor $x'$ out of $- 3 p x'$ .

$x' (2 x) + x' (- 3 p) = - 2 p + 3 x$

Factor $x'$ out of $x' (2 x) + x' (- 3 p)$ .

$x' (2 x - 3 p) = - 2 p + 3 x$

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

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

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

Simplify the left side.

Cancel the common factor of $2 x - 3 p$ .

Cancel the common factor.

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

Divide $x'$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

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

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

Factor $- 1$ out of $3 x$ .

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

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

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

Simplify the expression.

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

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

Move the negative in front of the fraction.

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

Step 6

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

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

Find dx/dp x^2-3xp+p^2=5

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