Calculus Examples

$f (y) = y \ln (11 x + 5 y)$

Step 1

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 $f (y) = y$ and $g (y) = \ln (11 x + 5 y)$ .

$y \frac{d}{d y} [\ln (11 x + 5 y)] + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = \ln (y)$ and $g (y) = 11 x + 5 y$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $11 x + 5 y$ .

$y (\frac{d}{d u} [\ln (u)] \frac{d}{d y} [11 x + 5 y]) + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 2.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$y (\frac{1}{u} \frac{d}{d y} [11 x + 5 y]) + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 2.3

Replace all occurrences of $u$ with $11 x + 5 y$ .

$y (\frac{1}{11 x + 5 y} \frac{d}{d y} [11 x + 5 y]) + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 3

Differentiate.

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Step 3.1

Combine $\frac{1}{11 x + 5 y}$ and $y$ .

$\frac{y}{11 x + 5 y} \frac{d}{d y} [11 x + 5 y] + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 3.2

By the Sum Rule, the derivative of $11 x + 5 y$ with respect to $y$ is $\frac{d}{d y} [11 x] + \frac{d}{d y} [5 y]$ .

$\frac{y}{11 x + 5 y} (\frac{d}{d y} [11 x] + \frac{d}{d y} [5 y]) + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 3.3

Since $11 x$ is constant with respect to $y$ , the derivative of $11 x$ with respect to $y$ is $0$ .

$\frac{y}{11 x + 5 y} (0 + \frac{d}{d y} [5 y]) + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 3.4

Add $0$ and $\frac{d}{d y} [5 y]$ .

$\frac{y}{11 x + 5 y} \frac{d}{d y} [5 y] + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 3.5

Since $5$ is constant with respect to $y$ , the derivative of $5 y$ with respect to $y$ is $5 \frac{d}{d y} [y]$ .

$\frac{y}{11 x + 5 y} (5 \frac{d}{d y} [y]) + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 3.6

Combine $5$ and $\frac{y}{11 x + 5 y}$ .

$\frac{5 y}{11 x + 5 y} \frac{d}{d y} [y] + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 3.7

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

$\frac{5 y}{11 x + 5 y} \cdot 1 + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 3.8

Multiply $\frac{5 y}{11 x + 5 y}$ by $1$ .

$\frac{5 y}{11 x + 5 y} + \ln (11 x + 5 y) \frac{d}{d y} [y]$

Step 3.9

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

$\frac{5 y}{11 x + 5 y} + \ln (11 x + 5 y) \cdot 1$

Step 3.10

Multiply $\ln (11 x + 5 y)$ by $1$ .

$\frac{5 y}{11 x + 5 y} + \ln (11 x + 5 y)$

Step 4

To write $\ln (11 x + 5 y)$ as a fraction with a common denominator, multiply by $\frac{11 x + 5 y}{11 x + 5 y}$ .

$\frac{5 y}{11 x + 5 y} + \frac{\ln (11 x + 5 y) (11 x + 5 y)}{11 x + 5 y}$

Step 5

Combine the numerators over the common denominator.

$\frac{5 y + \ln (11 x + 5 y) (11 x + 5 y)}{11 x + 5 y}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Simplify each term.

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Step 6.1.1.1

Apply the distributive property.

$\frac{5 y + \ln (11 x + 5 y) (11 x) + \ln (11 x + 5 y) (5 y)}{11 x + 5 y}$

Step 6.1.1.2

Rewrite using the commutative property of multiplication.

$\frac{5 y + 11 \ln (11 x + 5 y) x + \ln (11 x + 5 y) (5 y)}{11 x + 5 y}$

Step 6.1.1.3

Rewrite using the commutative property of multiplication.

$\frac{5 y + 11 \ln (11 x + 5 y) x + 5 \ln (11 x + 5 y) y}{11 x + 5 y}$

Step 6.1.1.4

Simplify each term.

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Step 6.1.1.4.1

Simplify $11 \ln (11 x + 5 y)$ by moving $11$ inside the logarithm.

$\frac{5 y + \ln ({(11 x + 5 y)}^{11}) x + 5 \ln (11 x + 5 y) y}{11 x + 5 y}$

Step 6.1.1.4.2

Simplify $5 \ln (11 x + 5 y)$ by moving $5$ inside the logarithm.

$\frac{5 y + \ln ({(11 x + 5 y)}^{11}) x + \ln ({(11 x + 5 y)}^{5}) y}{11 x + 5 y}$

Step 6.1.2

Reorder factors in $5 y + \ln ({(11 x + 5 y)}^{11}) x + \ln ({(11 x + 5 y)}^{5}) y$ .

$\frac{5 y + x \ln ({(11 x + 5 y)}^{11}) + y \ln ({(11 x + 5 y)}^{5})}{11 x + 5 y}$

Step 6.2

Reorder terms.

$\frac{x \ln ({(11 x + 5 y)}^{11}) + y \ln ({(11 x + 5 y)}^{5}) + 5 y}{11 x + 5 y}$