Calculus Examples

$y \cdot 2^{y} = x$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y \cdot 2^{y}) = \frac{d}{d x} (x)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 2^{x}$ and $g (x) = y$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $y$ .

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

Step 2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $2$ .

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

Step 2.2.3

Replace all occurrences of $u$ with $y$ .

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

Step 2.3

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

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

Step 2.4

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

$y (2^{y} \ln (2) y') + 2^{y} y'$

Step 2.5

Simplify.

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Step 2.5.1

Reorder terms.

$2^{y} y \ln (2) y' + 2^{y} y'$

Step 2.5.2

Reorder factors in $2^{y} y \ln (2) y' + 2^{y} y'$ .

$y \cdot 2^{y} \ln (2) y' + 2^{y} y'$

Step 3

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

$1$

Step 4

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

$y \cdot (2^{y} \ln (2) y') + 2^{y} y' = 1$

Step 5

Solve for $y'$ .

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Step 5.1

Simplify $y \cdot (2^{y} \ln (2) y') + 2^{y} y'$ .

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Step 5.1.1

Rewrite using the commutative property of multiplication.

$\ln (2) y (2^{y} y') + 2^{y} y' = 1$

Step 5.1.2

Reorder factors in $\ln (2) y (2^{y} y') + 2^{y} y'$ .

$y y' \cdot 2^{y} \ln (2) + y' \cdot 2^{y} = 1$

Step 5.2

Factor $y' \cdot 2^{y}$ out of $y y' \cdot (2^{y} \ln (2)) + y' \cdot 2^{y}$ .

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Step 5.2.1

Factor $y' \cdot 2^{y}$ out of $y y' \cdot (2^{y} \ln (2))$ .

$y' \cdot (2^{y} (y \cdot \ln (2))) + y' \cdot 2^{y} = 1$

Step 5.2.2

Factor $y' \cdot 2^{y}$ out of $y' \cdot 2^{y}$ .

$y' \cdot (2^{y} (y \cdot \ln (2))) + y' \cdot (2^{y} (1)) = 1$

Step 5.2.3

Factor $y' \cdot 2^{y}$ out of $y' \cdot 2^{y} (y \cdot \ln (2)) + y' \cdot 2^{y} \cdot 1$ .

$y' \cdot (2^{y} (y \cdot \ln (2) + 1)) = 1$

Step 5.3

Multiply $y$ by $\ln (2)$ .

$y' \cdot (2^{y} (y (\ln (2)) + 1)) = 1$

Step 5.4

Multiply $y$ by $\ln (2)$ .

$y' \cdot (2^{y} (y \ln (2) + 1)) = 1$

Step 5.5

Divide each term in $y' \cdot (2^{y} (y \ln (2) + 1)) = 1$ by $2^{y} (y \ln (2) + 1)$ and simplify.

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Step 5.5.1

Divide each term in $y' \cdot (2^{y} (y \ln (2) + 1)) = 1$ by $2^{y} (y \ln (2) + 1)$ .

$\frac{y' \cdot (2^{y} (y \ln (2) + 1))}{2^{y} (y \ln (2) + 1)} = \frac{1}{2^{y} (y \ln (2) + 1)}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

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

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Step 5.5.2.1.1

Cancel the common factor.

$\frac{y' \cdot (2^{y} (y \ln (2) + 1))}{2^{y} (y \ln (2) + 1)} = \frac{1}{2^{y} (y \ln (2) + 1)}$

Step 5.5.2.1.2

Rewrite the expression.

$\frac{y' \cdot (y \ln (2) + 1)}{y \ln (2) + 1} = \frac{1}{2^{y} (y \ln (2) + 1)}$

Step 5.5.2.2

Cancel the common factor of $y \ln (2) + 1$ .

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Step 5.5.2.2.1

Cancel the common factor.

$\frac{y' \cdot (y \ln (2) + 1)}{y \ln (2) + 1} = \frac{1}{2^{y} (y \ln (2) + 1)}$

Step 5.5.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{1}{2^{y} (y \ln (2) + 1)}$

Step 6

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

$\frac{d y}{d x} = \frac{1}{2^{y} (y \ln (2) + 1)}$