Find dx/dy y=xe^(3x-1)

$y = x e^{3 x - 1}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (y) = \frac{d}{d y} (x e^{3 x - 1})$

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.

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 $g (y) = e^{3 x - 1}$ .

$x \frac{d}{d y} [e^{3 x - 1}] + e^{3 x - 1} \frac{d}{d y} [x]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = e^{y}$ and $g (y) = 3 x - 1$ .

To apply the Chain Rule, set $u$ as $3 x - 1$ .

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

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

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

Replace all occurrences of $u$ with $3 x - 1$ .

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

Differentiate.

By the Sum Rule, the derivative of $3 x - 1$ with respect to $y$ is $\frac{d}{d y} [3 x] + \frac{d}{d y} [- 1]$ .

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

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

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

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

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

Since $- 1$ is constant with respect to $y$ , the derivative of $- 1$ with respect to $y$ is $0$ .

$x (e^{3 x - 1} (3 x' + 0)) + e^{3 x - 1} \frac{d}{d y} [x]$

Add $3 x'$ and $0$ .

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

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

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

Simplify.

Reorder terms.

$3 e^{3 x - 1} x x' + e^{3 x - 1} x'$

Reorder factors in $3 e^{3 x - 1} x x' + e^{3 x - 1} x'$ .

$3 x e^{3 x - 1} x' + e^{3 x - 1} x'$

Step 4

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

$1 = 3 x e^{3 x - 1} x' + e^{3 x - 1} x'$

Step 5

Solve for $x'$ .

Rewrite the equation as $3 x e^{3 x - 1} x' + e^{3 x - 1} x' = 1$ .

$3 x e^{3 x - 1} x' + e^{3 x - 1} x' = 1$

Reorder factors in $3 x e^{3 x - 1} x' + e^{3 x - 1} x'$ .

$3 x x' e^{3 x - 1} + x' e^{3 x - 1} = 1$

Factor $x' e^{3 x - 1}$ out of $3 x x' e^{3 x - 1} + x' e^{3 x - 1}$ .

Factor $x' e^{3 x - 1}$ out of $3 x x' e^{3 x - 1}$ .

$x' e^{3 x - 1} (3 x) + x' e^{3 x - 1} = 1$

Factor $x' e^{3 x - 1}$ out of $x' e^{3 x - 1}$ .

$x' e^{3 x - 1} (3 x) + x' e^{3 x - 1} \cdot 1 = 1$

Factor $x' e^{3 x - 1}$ out of $x' e^{3 x - 1} (3 x) + x' e^{3 x - 1} \cdot 1$ .

$x' e^{3 x - 1} (3 x + 1) = 1$

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

Divide each term in $x' e^{3 x - 1} (3 x + 1) = 1$ by $e^{3 x - 1} (3 x + 1)$ .

$\frac{x' e^{3 x - 1} (3 x + 1)}{e^{3 x - 1} (3 x + 1)} = \frac{1}{e^{3 x - 1} (3 x + 1)}$

Simplify the left side.

Cancel the common factor of $e^{3 x - 1}$ .

Cancel the common factor.

$\frac{x' e^{3 x - 1} (3 x + 1)}{e^{3 x - 1} (3 x + 1)} = \frac{1}{e^{3 x - 1} (3 x + 1)}$

Rewrite the expression.

$\frac{x' (3 x + 1)}{3 x + 1} = \frac{1}{e^{3 x - 1} (3 x + 1)}$

Cancel the common factor of $3 x + 1$ .

Cancel the common factor.

$\frac{x' (3 x + 1)}{3 x + 1} = \frac{1}{e^{3 x - 1} (3 x + 1)}$

Divide $x'$ by $1$ .

$x' = \frac{1}{e^{3 x - 1} (3 x + 1)}$

Step 6

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

$\frac{d x}{d y} = \frac{1}{e^{3 x - 1} (3 x + 1)}$

Find dx/dy y=xe^(3x-1)

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