Calculus Examples

$z = (2 x + 3 y) e^{4 x + 5 y}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (z) = \frac{d}{d x} ((2 x + 3 y) e^{4 x + 5 y})$

Step 2

The derivative of $z$ with respect to $x$ is $z'$ .

$z'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 2 x + 3 y$ and $g (x) = e^{4 x + 5 y}$ .

$(2 x + 3 y) \frac{d}{d x} [e^{4 x + 5 y}] + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 4 x + 5 y$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $4 x + 5 y$ .

$(2 x + 3 y) (\frac{d}{d u} [e^{u}] \frac{d}{d x} [4 x + 5 y]) + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.2.2

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

$(2 x + 3 y) (e^{u} \frac{d}{d x} [4 x + 5 y]) + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.2.3

Replace all occurrences of $u$ with $4 x + 5 y$ .

$(2 x + 3 y) (e^{4 x + 5 y} \frac{d}{d x} [4 x + 5 y]) + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.3

Differentiate.

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Step 3.3.1

By the Sum Rule, the derivative of $4 x + 5 y$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [5 y]$ .

$(2 x + 3 y) e^{4 x + 5 y} (\frac{d}{d x} [4 x] + \frac{d}{d x} [5 y]) + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.3.2

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

$(2 x + 3 y) e^{4 x + 5 y} (4 \frac{d}{d x} [x] + \frac{d}{d x} [5 y]) + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.3.3

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

$(2 x + 3 y) e^{4 x + 5 y} (4 \cdot 1 + \frac{d}{d x} [5 y]) + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.3.4

Multiply $4$ by $1$ .

$(2 x + 3 y) e^{4 x + 5 y} (4 + \frac{d}{d x} [5 y]) + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.3.5

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

$(2 x + 3 y) e^{4 x + 5 y} (4 + 0) + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.3.6

Simplify the expression.

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Step 3.3.6.1

Add $4$ and $0$ .

$(2 x + 3 y) e^{4 x + 5 y} \cdot 4 + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.3.6.2

Move $4$ to the left of $(2 x + 3 y) e^{4 x + 5 y}$ .

$4 \cdot ((2 x + 3 y) e^{4 x + 5 y}) + e^{4 x + 5 y} \frac{d}{d x} [2 x + 3 y]$

Step 3.3.7

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

$4 (2 x + 3 y) e^{4 x + 5 y} + e^{4 x + 5 y} (\frac{d}{d x} [2 x] + \frac{d}{d x} [3 y])$

Step 3.3.8

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

$4 (2 x + 3 y) e^{4 x + 5 y} + e^{4 x + 5 y} (2 \frac{d}{d x} [x] + \frac{d}{d x} [3 y])$

Step 3.3.9

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

$4 (2 x + 3 y) e^{4 x + 5 y} + e^{4 x + 5 y} (2 \cdot 1 + \frac{d}{d x} [3 y])$

Step 3.3.10

Multiply $2$ by $1$ .

$4 (2 x + 3 y) e^{4 x + 5 y} + e^{4 x + 5 y} (2 + \frac{d}{d x} [3 y])$

Step 3.3.11

Since $3 y$ is constant with respect to $x$ , the derivative of $3 y$ with respect to $x$ is $0$ .

$4 (2 x + 3 y) e^{4 x + 5 y} + e^{4 x + 5 y} (2 + 0)$

Step 3.3.12

Simplify the expression.

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Step 3.3.12.1

Add $2$ and $0$ .

$4 (2 x + 3 y) e^{4 x + 5 y} + e^{4 x + 5 y} \cdot 2$

Step 3.3.12.2

Move $2$ to the left of $e^{4 x + 5 y}$ .

$4 (2 x + 3 y) e^{4 x + 5 y} + 2 \cdot e^{4 x + 5 y}$

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

$(4 (2 x) + 4 (3 y)) e^{4 x + 5 y} + 2 e^{4 x + 5 y}$

Step 3.4.2

Combine terms.

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Step 3.4.2.1

Multiply $2$ by $4$ .

$(8 x + 4 (3 y)) e^{4 x + 5 y} + 2 e^{4 x + 5 y}$

Step 3.4.2.2

Multiply $3$ by $4$ .

$(8 x + 12 y) e^{4 x + 5 y} + 2 e^{4 x + 5 y}$

Step 3.4.3

Reorder terms.

$e^{4 x + 5 y} (8 x + 12 y) + 2 e^{4 x + 5 y}$

Step 4

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

$z' = e^{4 x + 5 y} (8 x + 12 y) + 2 e^{4 x + 5 y}$

Step 5

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

$\frac{d z}{d x} = e^{4 x + 5 y} (8 x + 12 y) + 2 e^{4 x + 5 y}$