Calculus Examples

$e^{x y^{2}} + \ln (y + z)$

Step 1

By the Sum Rule, the derivative of $e^{x y^{2}} + \ln (y + z)$ with respect to $y$ is $\frac{d}{d y} [e^{x y^{2}}] + \frac{d}{d y} [\ln (y + z)]$ .

$\frac{d}{d y} [e^{x y^{2}}] + \frac{d}{d y} [\ln (y + z)]$

Step 2

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

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Step 2.1

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 $g (y) = x y^{2}$ .

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Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $x y^{2}$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d y} [x y^{2}] + \frac{d}{d y} [\ln (y + z)]$

Step 2.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d y} [x y^{2}] + \frac{d}{d y} [\ln (y + z)]$

Step 2.1.3

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

$e^{x y^{2}} \frac{d}{d y} [x y^{2}] + \frac{d}{d y} [\ln (y + z)]$

Step 2.2

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

$e^{x y^{2}} (x \frac{d}{d y} [y^{2}]) + \frac{d}{d y} [\ln (y + z)]$

Step 2.3

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

$e^{x y^{2}} (x (2 y)) + \frac{d}{d y} [\ln (y + z)]$

Step 2.4

Move $2$ to the left of $x$ .

$e^{x y^{2}} (2 x y) + \frac{d}{d y} [\ln (y + z)]$

Step 3

Evaluate $\frac{d}{d y} [\ln (y + z)]$ .

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Step 3.1

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 $g (y) = y + z$ .

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Step 3.1.1

To apply the Chain Rule, set $u_{2}$ as $y + z$ .

$e^{x y^{2}} (2 x y) + \frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d y} [y + z]$

Step 3.1.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$e^{x y^{2}} (2 x y) + \frac{1}{u_{2}} \frac{d}{d y} [y + z]$

Step 3.1.3

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

$e^{x y^{2}} (2 x y) + \frac{1}{y + z} \frac{d}{d y} [y + z]$

Step 3.2

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

$e^{x y^{2}} (2 x y) + \frac{1}{y + z} (\frac{d}{d y} [y] + \frac{d}{d y} [z])$

Step 3.3

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

$e^{x y^{2}} (2 x y) + \frac{1}{y + z} (1 + \frac{d}{d y} [z])$

Step 3.4

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

$e^{x y^{2}} (2 x y) + \frac{1}{y + z} (1 + 0)$

Step 3.5

Add $1$ and $0$ .

$e^{x y^{2}} (2 x y) + \frac{1}{y + z} \cdot 1$

Step 3.6

Multiply $\frac{1}{y + z}$ by $1$ .

$e^{x y^{2}} (2 x y) + \frac{1}{y + z}$

Step 4

Simplify.

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Step 4.1

Combine terms.

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Step 4.1.1

To write $e^{x y^{2}} (2 x y)$ as a fraction with a common denominator, multiply by $\frac{y + z}{y + z}$ .

$\frac{e^{x y^{2}} (2 x y) (y + z)}{y + z} + \frac{1}{y + z}$

Step 4.1.2

Combine the numerators over the common denominator.

$\frac{e^{x y^{2}} (2 x y) (y + z) + 1}{y + z}$

Step 4.2

Reorder terms.

$\frac{2 e^{x y^{2}} x y (y + z) + 1}{y + z}$