Find dx/dy y=x(x-4)^3

$y = x {(x - 4)}^{3}$

Step 1

Differentiate both sides of the equation.

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

Step 2

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

$1$

Step 3

Differentiate the right side of the equation.

Use the Binomial Theorem.

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

Simplify each term.

Multiply $- 4$ by $3$ .

$\frac{d}{d y} [x (x^{3} - 12 x^{2} + 3 x {(- 4)}^{2} + {(- 4)}^{3})]$

Raise $- 4$ to the power of $2$ .

$\frac{d}{d y} [x (x^{3} - 12 x^{2} + 3 x \cdot 16 + {(- 4)}^{3})]$

Multiply $16$ by $3$ .

$\frac{d}{d y} [x (x^{3} - 12 x^{2} + 48 x + {(- 4)}^{3})]$

Raise $- 4$ to the power of $3$ .

$\frac{d}{d y} [x (x^{3} - 12 x^{2} + 48 x - 64)]$

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = x$ and $g (y) = x^{3} - 12 x^{2} + 48 x - 64$ .

$x \frac{d}{d y} [x^{3} - 12 x^{2} + 48 x - 64] + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

By the Sum Rule, the derivative of $x^{3} - 12 x^{2} + 48 x - 64$ with respect to $y$ is $\frac{d}{d y} [x^{3}] + \frac{d}{d y} [- 12 x^{2}] + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]$ .

$x (\frac{d}{d y} [x^{3}] + \frac{d}{d y} [- 12 x^{2}] + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{3}$ and $g (y) = x$ .

To apply the Chain Rule, set $u_{1}$ as $x$ .

$x (\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d y} [x] + \frac{d}{d y} [- 12 x^{2}] + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

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

$x (3 {u_{1}}^{2} \frac{d}{d y} [x] + \frac{d}{d y} [- 12 x^{2}] + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

Replace all occurrences of $u_{1}$ with $x$ .

$x (3 x^{2} \frac{d}{d y} [x] + \frac{d}{d y} [- 12 x^{2}] + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

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

$x (3 x^{2} x' + \frac{d}{d y} [- 12 x^{2}] + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

Since $- 12$ is constant with respect to $y$ , the derivative of $- 12 x^{2}$ with respect to $y$ is $- 12 \frac{d}{d y} [x^{2}]$ .

$x (3 x^{2} x' - 12 \frac{d}{d y} [x^{2}] + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

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

To apply the Chain Rule, set $u_{2}$ as $x$ .

$x (3 x^{2} x' - 12 (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d y} [x]) + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

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

$x (3 x^{2} x' - 12 (2 u_{2} \frac{d}{d y} [x]) + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

Replace all occurrences of $u_{2}$ with $x$ .

$x (3 x^{2} x' - 12 (2 x \frac{d}{d y} [x]) + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

Multiply $2$ by $- 12$ .

$x (3 x^{2} x' - 24 (x \frac{d}{d y} [x]) + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

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

$x (3 x^{2} x' - 24 x x' + \frac{d}{d y} [48 x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

Since $48$ is constant with respect to $y$ , the derivative of $48 x$ with respect to $y$ is $48 \frac{d}{d y} [x]$ .

$x (3 x^{2} x' - 24 x x' + 48 \frac{d}{d y} [x] + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

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

$x (3 x^{2} x' - 24 x x' + 48 x' + \frac{d}{d y} [- 64]) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

Since $- 64$ is constant with respect to $y$ , the derivative of $- 64$ with respect to $y$ is $0$ .

$x (3 x^{2} x' - 24 x x' + 48 x' + 0) + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

Add $3 x^{2} x' - 24 x x' + 48 x'$ and $0$ .

$x (3 x^{2} x' - 24 x x' + 48 x') + (x^{3} - 12 x^{2} + 48 x - 64) \frac{d}{d y} [x]$

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

$x (3 x^{2} x' - 24 x x' + 48 x') + (x^{3} - 12 x^{2} + 48 x - 64) x'$

Simplify.

Apply the distributive property.

$x (3 x^{2} x') + x (- 24 x x') + x (48 x') + (x^{3} - 12 x^{2} + 48 x - 64) x'$

Apply the distributive property.

$x (3 x^{2} x') + x (- 24 x x') + x (48 x') + x^{3} x' - 12 x^{2} x' + 48 x x' - 64 x'$

Combine terms.

Raise $x$ to the power of $1$ .

$3 (x^{1} x^{2}) x' + x (- 24 x x') + x (48 x') + x^{3} x' - 12 x^{2} x' + 48 x x' - 64 x'$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3 x^{1 + 2} x' + x (- 24 x x') + x (48 x') + x^{3} x' - 12 x^{2} x' + 48 x x' - 64 x'$

Add $1$ and $2$ .

$3 x^{3} x' + x (- 24 x x') + x (48 x') + x^{3} x' - 12 x^{2} x' + 48 x x' - 64 x'$

Raise $x$ to the power of $1$ .

$3 x^{3} x' - 24 (x^{1} x) x' + x (48 x') + x^{3} x' - 12 x^{2} x' + 48 x x' - 64 x'$

Raise $x$ to the power of $1$ .

$3 x^{3} x' - 24 (x^{1} x^{1}) x' + x (48 x') + x^{3} x' - 12 x^{2} x' + 48 x x' - 64 x'$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3 x^{3} x' - 24 x^{1 + 1} x' + x (48 x') + x^{3} x' - 12 x^{2} x' + 48 x x' - 64 x'$

Add $1$ and $1$ .

$3 x^{3} x' - 24 x^{2} x' + x (48 x') + x^{3} x' - 12 x^{2} x' + 48 x x' - 64 x'$

Add $3 x^{3} x'$ and $x^{3} x'$ .

$4 x^{3} x' - 24 x^{2} x' + x (48 x') - 12 x^{2} x' + 48 x x' - 64 x'$

Subtract $12 x^{2} x'$ from $- 24 x^{2} x'$ .

$4 x^{3} x' - 36 x^{2} x' + x (48 x') + 48 x x' - 64 x'$

Add $x (48 x')$ and $48 x x'$ .

Reorder $x$ and $48$ .

$4 x^{3} x' - 36 x^{2} x' + 48 \cdot x x' + 48 x x' - 64 x'$

Add $48 \cdot x x'$ and $48 x x'$ .

$4 x^{3} x' - 36 x^{2} x' + 96 x x' - 64 x'$

Step 4

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

$1 = 4 x^{3} x' - 36 x^{2} x' + 96 x x' - 64 x'$

Step 5

Solve for $x'$ .

Rewrite the equation as $4 x^{3} x' - 36 x^{2} x' + 96 x x' - 64 x' = 1$ .

$4 x^{3} x' - 36 x^{2} x' + 96 x x' - 64 x' = 1$

Factor $4 x'$ out of $4 x^{3} x' - 36 x^{2} x' + 96 x x' - 64 x'$ .

Factor $4 x'$ out of $4 x^{3} x'$ .

$4 x' x^{3} - 36 x^{2} x' + 96 x x' - 64 x' = 1$

Factor $4 x'$ out of $- 36 x^{2} x'$ .

$4 x' x^{3} + 4 x' (- 9 x^{2}) + 96 x x' - 64 x' = 1$

Factor $4 x'$ out of $96 x x'$ .

$4 x' x^{3} + 4 x' (- 9 x^{2}) + 4 x' (24 x) - 64 x' = 1$

Factor $4 x'$ out of $- 64 x'$ .

$4 x' x^{3} + 4 x' (- 9 x^{2}) + 4 x' (24 x) + 4 x' \cdot - 16 = 1$

Factor $4 x'$ out of $4 x' x^{3} + 4 x' (- 9 x^{2})$ .

$4 x' (x^{3} - 9 x^{2}) + 4 x' (24 x) + 4 x' \cdot - 16 = 1$

Factor $4 x'$ out of $4 x' (x^{3} - 9 x^{2}) + 4 x' (24 x)$ .

$4 x' (x^{3} - 9 x^{2} + 24 x) + 4 x' \cdot - 16 = 1$

Factor $4 x'$ out of $4 x' (x^{3} - 9 x^{2} + 24 x) + 4 x' \cdot - 16$ .

$4 x' (x^{3} - 9 x^{2} + 24 x - 16) = 1$

Divide each term in $4 x' (x^{3} - 9 x^{2} + 24 x - 16) = 1$ by $4 (x^{3} - 9 x^{2} + 24 x - 16)$ and simplify.

Divide each term in $4 x' (x^{3} - 9 x^{2} + 24 x - 16) = 1$ by $4 (x^{3} - 9 x^{2} + 24 x - 16)$ .

$\frac{4 x' (x^{3} - 9 x^{2} + 24 x - 16)}{4 (x^{3} - 9 x^{2} + 24 x - 16)} = \frac{1}{4 (x^{3} - 9 x^{2} + 24 x - 16)}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x' (x^{3} - 9 x^{2} + 24 x - 16)}{4 (x^{3} - 9 x^{2} + 24 x - 16)} = \frac{1}{4 (x^{3} - 9 x^{2} + 24 x - 16)}$

Rewrite the expression.

$\frac{x' (x^{3} - 9 x^{2} + 24 x - 16)}{x^{3} - 9 x^{2} + 24 x - 16} = \frac{1}{4 (x^{3} - 9 x^{2} + 24 x - 16)}$

Cancel the common factor of $x^{3} - 9 x^{2} + 24 x - 16$ .

Cancel the common factor.

$\frac{x' (x^{3} - 9 x^{2} + 24 x - 16)}{x^{3} - 9 x^{2} + 24 x - 16} = \frac{1}{4 (x^{3} - 9 x^{2} + 24 x - 16)}$

Divide $x'$ by $1$ .

$x' = \frac{1}{4 (x^{3} - 9 x^{2} + 24 x - 16)}$

Step 6

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

$\frac{d x}{d y} = \frac{1}{4 (x^{3} - 9 x^{2} + 24 x - 16)}$

Find dx/dy y=x(x-4)^3

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