Calculus Examples

$\frac{d y}{d x} = \frac{10}{{(2 x - 1)}^{2} e^{y + 2}}$

Step 1

Separate the variables.

Tap for more steps...

Step 1.1

Regroup factors.

$\frac{d y}{d x} = \frac{10}{{(2 x - 1)}^{2}} \cdot \frac{1}{e^{y + 2}}$

Step 1.2

Multiply both sides by $e^{y + 2}$ .

$e^{y + 2} \frac{d y}{d x} = e^{y + 2} (\frac{10}{{(2 x - 1)}^{2}} \cdot \frac{1}{e^{y + 2}})$

Step 1.3

Simplify.

Tap for more steps...

Step 1.3.1

Combine.

$e^{y + 2} \frac{d y}{d x} = e^{y + 2} \frac{10 \cdot 1}{{(2 x - 1)}^{2} e^{y + 2}}$

Step 1.3.2

Cancel the common factor of $e^{y + 2}$ .

Tap for more steps...

Step 1.3.2.1

Factor $e^{y + 2}$ out of ${(2 x - 1)}^{2} e^{y + 2}$ .

$e^{y + 2} \frac{d y}{d x} = e^{y + 2} \frac{10 \cdot 1}{e^{y + 2} {(2 x - 1)}^{2}}$

Step 1.3.2.2

Cancel the common factor.

$e^{y + 2} \frac{d y}{d x} = e^{y + 2} \frac{10 \cdot 1}{e^{y + 2} {(2 x - 1)}^{2}}$

Step 1.3.2.3

Rewrite the expression.

$e^{y + 2} \frac{d y}{d x} = \frac{10 \cdot 1}{{(2 x - 1)}^{2}}$

Step 1.3.3

Multiply $10$ by $1$ .

$e^{y + 2} \frac{d y}{d x} = \frac{10}{{(2 x - 1)}^{2}}$

Step 1.4

Rewrite the equation.

$e^{y + 2} d y = \frac{10}{{(2 x - 1)}^{2}} d x$

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

$\int e^{y + 2} d y = \int \frac{10}{{(2 x - 1)}^{2}} d x$

Step 2.2

Integrate the left side.

Tap for more steps...

Step 2.2.1

Let $u_{1} = y + 2$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

Tap for more steps...

Step 2.2.1.1

Let $u_{1} = y + 2$ . Find $\frac{d u_{1}}{d y}$ .

Tap for more steps...

Step 2.2.1.1.1

Differentiate $y + 2$ .

$\frac{d}{d y} [y + 2]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $y + 2$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [2]$ .

$\frac{d}{d y} [y] + \frac{d}{d y} [2]$

Step 2.2.1.1.3

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

$1 + \frac{d}{d y} [2]$

Step 2.2.1.1.4

Since $2$ is constant with respect to $y$ , the derivative of $2$ with respect to $y$ is $0$ .

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int e^{u_{1}} d u_{1} = \int \frac{10}{{(2 x - 1)}^{2}} d x$

Step 2.2.2

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

$e^{u_{1}} + C_{1} = \int \frac{10}{{(2 x - 1)}^{2}} d x$

Step 2.2.3

Replace all occurrences of $u_{1}$ with $y + 2$ .

$e^{y + 2} + C_{1} = \int \frac{10}{{(2 x - 1)}^{2}} d x$

Step 2.3

Integrate the right side.

Tap for more steps...

Step 2.3.1

Since $10$ is constant with respect to $x$ , move $10$ out of the integral.

$e^{y + 2} + C_{1} = 10 \int \frac{1}{{(2 x - 1)}^{2}} d x$

Step 2.3.2

Let $u_{2} = 2 x - 1$ . Then $d u_{2} = 2 d x$ , so $\frac{1}{2} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Tap for more steps...

Step 2.3.2.1

Let $u_{2} = 2 x - 1$ . Find $\frac{d u_{2}}{d x}$ .

Tap for more steps...

Step 2.3.2.1.1

Differentiate $2 x - 1$ .

$\frac{d}{d x} [2 x - 1]$

Step 2.3.2.1.2

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$

Step 2.3.2.1.3

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

Tap for more steps...

Step 2.3.2.1.3.1

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

$2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 2.3.2.1.3.2

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

$2 \cdot 1 + \frac{d}{d x} [- 1]$

Step 2.3.2.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [- 1]$

Step 2.3.2.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 2.3.2.1.4.1

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

$2 + 0$

Step 2.3.2.1.4.2

Add $2$ and $0$ .

$2$

Step 2.3.2.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$e^{y + 2} + C_{1} = 10 \int \frac{1}{{u_{2}}^{2}} \cdot \frac{1}{2} d u_{2}$

Step 2.3.3

Simplify.

Tap for more steps...

Step 2.3.3.1

Multiply $\frac{1}{{u_{2}}^{2}}$ by $\frac{1}{2}$ .

$e^{y + 2} + C_{1} = 10 \int \frac{1}{{u_{2}}^{2} \cdot 2} d u_{2}$

Step 2.3.3.2

Move $2$ to the left of ${u_{2}}^{2}$ .

$e^{y + 2} + C_{1} = 10 \int \frac{1}{2 {u_{2}}^{2}} d u_{2}$

Step 2.3.4

Since $\frac{1}{2}$ is constant with respect to $u_{2}$ , move $\frac{1}{2}$ out of the integral.

$e^{y + 2} + C_{1} = 10 (\frac{1}{2} \int \frac{1}{{u_{2}}^{2}} d u_{2})$

Step 2.3.5

Simplify the expression.

Tap for more steps...

Step 2.3.5.1

Simplify.

Tap for more steps...

Step 2.3.5.1.1

Combine $\frac{1}{2}$ and $10$ .

$e^{y + 2} + C_{1} = \frac{10}{2} \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.1.2

Cancel the common factor of $10$ and $2$ .

Tap for more steps...

Step 2.3.5.1.2.1

Factor $2$ out of $10$ .

$e^{y + 2} + C_{1} = \frac{2 \cdot 5}{2} \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.1.2.2

Cancel the common factors.

Tap for more steps...

Step 2.3.5.1.2.2.1

Factor $2$ out of $2$ .

$e^{y + 2} + C_{1} = \frac{2 \cdot 5}{2 (1)} \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.1.2.2.2

Cancel the common factor.

$e^{y + 2} + C_{1} = \frac{2 \cdot 5}{2 \cdot 1} \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.1.2.2.3

Rewrite the expression.

$e^{y + 2} + C_{1} = \frac{5}{1} \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.1.2.2.4

Divide $5$ by $1$ .

$e^{y + 2} + C_{1} = 5 \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.2

Apply basic rules of exponents.

Tap for more steps...

Step 2.3.5.2.1

Move ${u_{2}}^{2}$ out of the denominator by raising it to the $- 1$ power.

$e^{y + 2} + C_{1} = 5 \int {({u_{2}}^{2})}^{- 1} d u_{2}$

Step 2.3.5.2.2

Multiply the exponents in ${({u_{2}}^{2})}^{- 1}$ .

Tap for more steps...

Step 2.3.5.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$e^{y + 2} + C_{1} = 5 \int {u_{2}}^{2 \cdot - 1} d u_{2}$

Step 2.3.5.2.2.2

Multiply $2$ by $- 1$ .

$e^{y + 2} + C_{1} = 5 \int {u_{2}}^{- 2} d u_{2}$

Step 2.3.6

By the Power Rule, the integral of ${u_{2}}^{- 2}$ with respect to $u_{2}$ is $- {u_{2}}^{- 1}$ .

$e^{y + 2} + C_{1} = 5 (- {u_{2}}^{- 1} + C_{2})$

Step 2.3.7

Simplify.

Tap for more steps...

Step 2.3.7.1

Rewrite $5 (- {u_{2}}^{- 1} + C_{2})$ as $5 (- \frac{1}{u_{2}}) + C_{2}$ .

$e^{y + 2} + C_{1} = 5 (- \frac{1}{u_{2}}) + C_{2}$

Step 2.3.7.2

Simplify.

Tap for more steps...

Step 2.3.7.2.1

Multiply $- 1$ by $5$ .

$e^{y + 2} + C_{1} = - 5 \frac{1}{u_{2}} + C_{2}$

Step 2.3.7.2.2

Combine $- 5$ and $\frac{1}{u_{2}}$ .

$e^{y + 2} + C_{1} = \frac{- 5}{u_{2}} + C_{2}$

Step 2.3.7.2.3

Move the negative in front of the fraction.

$e^{y + 2} + C_{1} = - \frac{5}{u_{2}} + C_{2}$

Step 2.3.8

Replace all occurrences of $u_{2}$ with $2 x - 1$ .

$e^{y + 2} + C_{1} = - \frac{5}{2 x - 1} + C_{2}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$e^{y + 2} = - \frac{5}{2 x - 1} + K$

Step 3

Solve for $y$ .

Tap for more steps...

Step 3.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{y + 2}) = \ln (\frac{- 5 + 2 K x - K}{2 x - 1})$

Step 3.2

Expand the left side.

Tap for more steps...

Step 3.2.1

Expand $\ln (e^{y + 2})$ by moving $y + 2$ outside the logarithm.

$(y + 2) \ln (e) = \ln (\frac{- 5 + 2 K x - K}{2 x - 1})$

Step 3.2.2

The natural logarithm of $e$ is $1$ .

$(y + 2) \cdot 1 = \ln (\frac{- 5 + 2 K x - K}{2 x - 1})$

Step 3.2.3

Multiply $y + 2$ by $1$ .

$y + 2 = \ln (\frac{- 5 + 2 K x - K}{2 x - 1})$

Step 3.3

Simplify the right side.

Tap for more steps...

Step 3.3.1

Simplify $\ln (\frac{- 5 + 2 K x - K}{2 x - 1})$ .

Tap for more steps...

Step 3.3.1.1

Split the fraction $\frac{- 5 + 2 K x - K}{2 x - 1}$ into two fractions.

$y + 2 = \ln (\frac{- 5 + 2 K x}{2 x - 1} + \frac{- K}{2 x - 1})$

Step 3.3.1.2

Simplify each term.

Tap for more steps...

Step 3.3.1.2.1

Split the fraction $\frac{- 5 + 2 K x}{2 x - 1}$ into two fractions.

$y + 2 = \ln (\frac{- 5}{2 x - 1} + \frac{2 K x}{2 x - 1} + \frac{- K}{2 x - 1})$

Step 3.3.1.2.2

Move the negative in front of the fraction.

$y + 2 = \ln (- \frac{5}{2 x - 1} + \frac{2 K x}{2 x - 1} + \frac{- K}{2 x - 1})$

Step 3.3.1.2.3

Move the negative in front of the fraction.

$y + 2 = \ln (- \frac{5}{2 x - 1} + \frac{2 K x}{2 x - 1} - \frac{K}{2 x - 1})$

Step 3.4

Subtract $2$ from both sides of the equation.

$y = \ln (- \frac{5}{2 x - 1} + \frac{2 K x}{2 x - 1} - \frac{K}{2 x - 1}) - 2$

Step 4

Simplify the constant of integration.

$y = \ln (- \frac{5}{2 x - 1} + \frac{K x}{2 x - 1} - \frac{K}{2 x - 1}) - 2$