Calculus Examples

$y + \cos (y) = x + 1$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y + \cos (y)) = \frac{d}{d x} (x + 1)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

$\frac{d}{d x} [y] + \frac{d}{d x} [\cos (y)]$

Step 2.2

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

$y' + \frac{d}{d x} [\cos (y)]$

Step 2.3

Evaluate $\frac{d}{d x} [\cos (y)]$ .

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Step 2.3.1

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

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Step 2.3.1.1

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

$y' + \frac{d}{d u} [\cos (u)] \frac{d}{d x} [y]$

Step 2.3.1.2

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$y' - \sin (u) \frac{d}{d x} [y]$

Step 2.3.1.3

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

$y' - \sin (y) \frac{d}{d x} [y]$

Step 2.3.2

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

$y' - \sin (y) y'$

Step 2.4

Reorder terms.

$- \sin (y) y' + y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

$1 + 0$

Step 3.4

Add $1$ and $0$ .

$1$

Step 4

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

$- \sin (y) y' + y' = 1$

Step 5

Solve for $y'$ .

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Step 5.1

Simplify the left side.

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Step 5.1.1

Reorder factors in $- \sin (y) y' + y'$ .

$- y' \sin (y) + y' = 1$

Step 5.2

Factor $y'$ out of $- y' \sin (y) + y'$ .

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Step 5.2.1

Factor $y'$ out of $- y' \sin (y)$ .

$y' (- 1 \sin (y)) + y' = 1$

Step 5.2.2

Raise $y'$ to the power of $1$ .

$y' (- 1 \sin (y)) + y' = 1$

Step 5.2.3

Factor $y'$ out of ${y'}^{1}$ .

$y' (- 1 \sin (y)) + y' \cdot 1 = 1$

Step 5.2.4

Factor $y'$ out of $y' (- 1 \sin (y)) + y' \cdot 1$ .

$y' (- 1 \sin (y) + 1) = 1$

Step 5.3

Rewrite $- 1 \sin (y)$ as $- \sin (y)$ .

$y' (- \sin (y) + 1) = 1$

Step 5.4

Divide each term in $y' (- \sin (y) + 1) = 1$ by $- \sin (y) + 1$ and simplify.

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Step 5.4.1

Divide each term in $y' (- \sin (y) + 1) = 1$ by $- \sin (y) + 1$ .

$\frac{y' (- \sin (y) + 1)}{- \sin (y) + 1} = \frac{1}{- \sin (y) + 1}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $- \sin (y) + 1$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{y' (- \sin (y) + 1)}{- \sin (y) + 1} = \frac{1}{- \sin (y) + 1}$

Step 5.4.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{1}{- \sin (y) + 1}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Factor $- 1$ out of $- \sin (y)$ .

$y' = \frac{1}{- (\sin (y)) + 1}$

Step 5.4.3.2

Rewrite $1$ as $- 1 (- 1)$ .

$y' = \frac{1}{- (\sin (y)) - 1 (- 1)}$

Step 5.4.3.3

Factor $- 1$ out of $- (\sin (y)) - 1 (- 1)$ .

$y' = \frac{1}{- (\sin (y) - 1)}$

Step 5.4.3.4

Rewrite negatives.

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Step 5.4.3.4.1

Rewrite $- (\sin (y) - 1)$ as $- 1 (\sin (y) - 1)$ .

$y' = \frac{1}{- 1 (\sin (y) - 1)}$

Step 5.4.3.4.2

Move the negative in front of the fraction.

$y' = - \frac{1}{\sin (y) - 1}$

Step 6

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

$\frac{d y}{d x} = - \frac{1}{\sin (y) - 1}$

Step 7

Set $\frac{d y}{d x} = 0$ then solve for $x$ in terms of $y$ .

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Step 7.1

Set the numerator equal to zero.

$1 = 0$

Step 7.2

Since $1 \neq 0$ , there are no solutions.

No solution

Step 8

No points that set $\frac{d y}{d x} = 0$ are on the real number plane.

No Points

Step 9