Find dx/dy y=3x^3+6x^2-4x

$y = 3 x^{3} + 6 x^{2} - 4 x$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (y) = \frac{d}{d y} (3 x^{3} + 6 x^{2} - 4 x)$

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.

By the Sum Rule, the derivative of $3 x^{3} + 6 x^{2} - 4 x$ with respect to $y$ is $\frac{d}{d y} [3 x^{3}] + \frac{d}{d y} [6 x^{2}] + \frac{d}{d y} [- 4 x]$ .

$\frac{d}{d y} [3 x^{3}] + \frac{d}{d y} [6 x^{2}] + \frac{d}{d y} [- 4 x]$

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

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

$3 \frac{d}{d y} [x^{3}] + \frac{d}{d y} [6 x^{2}] + \frac{d}{d y} [- 4 x]$

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 $x$ .

$3 (\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d y} [x]) + \frac{d}{d y} [6 x^{2}] + \frac{d}{d y} [- 4 x]$

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

$3 (3 {u_{1}}^{2} \frac{d}{d y} [x]) + \frac{d}{d y} [6 x^{2}] + \frac{d}{d y} [- 4 x]$

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

$3 (3 x^{2} \frac{d}{d y} [x]) + \frac{d}{d y} [6 x^{2}] + \frac{d}{d y} [- 4 x]$

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

$3 (3 x^{2} x') + \frac{d}{d y} [6 x^{2}] + \frac{d}{d y} [- 4 x]$

Multiply $3$ by $3$ .

$9 x^{2} x' + \frac{d}{d y} [6 x^{2}] + \frac{d}{d y} [- 4 x]$

Evaluate $\frac{d}{d y} [6 x^{2}]$ .

Since $6$ is constant with respect to $y$ , the derivative of $6 x^{2}$ with respect to $y$ is $6 \frac{d}{d y} [x^{2}]$ .

$9 x^{2} x' + 6 \frac{d}{d y} [x^{2}] + \frac{d}{d y} [- 4 x]$

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 $x$ .

$9 x^{2} x' + 6 (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d y} [x]) + \frac{d}{d y} [- 4 x]$

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

$9 x^{2} x' + 6 (2 u_{2} \frac{d}{d y} [x]) + \frac{d}{d y} [- 4 x]$

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

$9 x^{2} x' + 6 (2 x \frac{d}{d y} [x]) + \frac{d}{d y} [- 4 x]$

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

$9 x^{2} x' + 6 (2 x x') + \frac{d}{d y} [- 4 x]$

Multiply $2$ by $6$ .

$9 x^{2} x' + 12 x x' + \frac{d}{d y} [- 4 x]$

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

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

$9 x^{2} x' + 12 x x' - 4 \frac{d}{d y} [x]$

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

$9 x^{2} x' + 12 x x' - 4 x'$

Step 4

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

$1 = 9 x^{2} x' + 12 x x' - 4 x'$

Step 5

Solve for $x'$ .

Rewrite the equation as $9 x^{2} x' + 12 x x' - 4 x' = 1$ .

$9 x^{2} x' + 12 x x' - 4 x' = 1$

Factor $x'$ out of $9 x^{2} x' + 12 x x' - 4 x'$ .

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

$x' (9 x^{2}) + 12 x x' - 4 x' = 1$

Factor $x'$ out of $12 x x'$ .

$x' (9 x^{2}) + x' (12 x) - 4 x' = 1$

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

$x' (9 x^{2}) + x' (12 x) + x' \cdot - 4 = 1$

Factor $x'$ out of $x' (9 x^{2}) + x' (12 x)$ .

$x' (9 x^{2} + 12 x) + x' \cdot - 4 = 1$

Factor $x'$ out of $x' (9 x^{2} + 12 x) + x' \cdot - 4$ .

$x' (9 x^{2} + 12 x - 4) = 1$

Divide each term in $x' (9 x^{2} + 12 x - 4) = 1$ by $9 x^{2} + 12 x - 4$ and simplify.

Divide each term in $x' (9 x^{2} + 12 x - 4) = 1$ by $9 x^{2} + 12 x - 4$ .

$\frac{x' (9 x^{2} + 12 x - 4)}{9 x^{2} + 12 x - 4} = \frac{1}{9 x^{2} + 12 x - 4}$

Simplify the left side.

Cancel the common factor of $9 x^{2} + 12 x - 4$ .

Cancel the common factor.

$\frac{x' (9 x^{2} + 12 x - 4)}{9 x^{2} + 12 x - 4} = \frac{1}{9 x^{2} + 12 x - 4}$

Divide $x'$ by $1$ .

$x' = \frac{1}{9 x^{2} + 12 x - 4}$

Step 6

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

$\frac{d x}{d y} = \frac{1}{9 x^{2} + 12 x - 4}$

Find dx/dy y=3x^3+6x^2-4x

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