Calculus Examples

$y = \sin (x) \cos (x)$

Step 1

Differentiate both sides of the equation.

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

Step 2

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

$1$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = \sin (x)$ and $g (y) = \cos (x)$ .

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

Step 3.2

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

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Step 3.2.1

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

$\sin (x) (\frac{d}{d u_{1}} [\cos (u_{1})] \frac{d}{d y} [x]) + \cos (x) \frac{d}{d y} [\sin (x)]$

Step 3.2.2

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

$\sin (x) (- \sin (u_{1}) \frac{d}{d y} [x]) + \cos (x) \frac{d}{d y} [\sin (x)]$

Step 3.2.3

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

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

Step 3.3

Raise $\sin (x)$ to the power of $1$ .

$- (\sin^{1} (x) \sin (x)) \frac{d}{d y} [x] + \cos (x) \frac{d}{d y} [\sin (x)]$

Step 3.4

Raise $\sin (x)$ to the power of $1$ .

$- (\sin^{1} (x) \sin^{1} (x)) \frac{d}{d y} [x] + \cos (x) \frac{d}{d y} [\sin (x)]$

Step 3.5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- {\sin (x)}^{1 + 1} \frac{d}{d y} [x] + \cos (x) \frac{d}{d y} [\sin (x)]$

Step 3.6

Add $1$ and $1$ .

$- \sin^{2} (x) \frac{d}{d y} [x] + \cos (x) \frac{d}{d y} [\sin (x)]$

Step 3.7

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

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

Step 3.8

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

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Step 3.8.1

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

$- \sin^{2} (x) x' + \cos (x) (\frac{d}{d u_{2}} [\sin (u_{2})] \frac{d}{d y} [x])$

Step 3.8.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

$- \sin^{2} (x) x' + \cos (x) (\cos (u_{2}) \frac{d}{d y} [x])$

Step 3.8.3

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

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

Step 3.9

Raise $\cos (x)$ to the power of $1$ .

$- \sin^{2} (x) x' + \cos^{1} (x) \cos (x) \frac{d}{d y} [x]$

Step 3.10

Raise $\cos (x)$ to the power of $1$ .

$- \sin^{2} (x) x' + \cos^{1} (x) \cos^{1} (x) \frac{d}{d y} [x]$

Step 3.11

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \sin^{2} (x) x' + {\cos (x)}^{1 + 1} \frac{d}{d y} [x]$

Step 3.12

Add $1$ and $1$ .

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

Step 3.13

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

$- \sin^{2} (x) x' + \cos^{2} (x) x'$

Step 4

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

$1 = - \sin^{2} (x) x' + \cos^{2} (x) x'$

Step 5

Solve for $x'$ .

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Step 5.1

Rewrite the equation as $- \sin^{2} (x) x' + \cos^{2} (x) x' = 1$ .

$- \sin^{2} (x) x' + \cos^{2} (x) x' = 1$

Step 5.2

Simplify the left side.

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Step 5.2.1

Reorder factors in $- \sin^{2} (x) x' + \cos^{2} (x) x'$ .

$- x' \sin^{2} (x) + x' \cos^{2} (x) = 1$

Step 5.3

Factor $x'$ out of $- x' \sin^{2} (x) + x' \cos^{2} (x)$ .

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Step 5.3.1

Factor $x'$ out of $- x' \sin^{2} (x)$ .

$x' (- 1 \sin^{2} (x)) + x' (\cos^{2} (x)) = 1$

Step 5.3.2

Factor $x'$ out of $x' \cos^{2} (x)$ .

$x' (- 1 \sin^{2} (x)) + x' (\cos^{2} (x)) = 1$

Step 5.3.3

Factor $x'$ out of $x' (- 1 \sin^{2} (x)) + x' (\cos^{2} (x))$ .

$x' (- 1 \sin^{2} (x) + \cos^{2} (x)) = 1$

Step 5.4

Reorder $- 1 \sin^{2} (x)$ and $\cos^{2} (x)$ .

$x' (\cos^{2} (x) - 1 \sin^{2} (x)) = 1$

Step 5.5

Factor.

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Step 5.5.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \cos (x)$ and $b = \sin (x)$ .

$x' ((\cos (x) + \sin (x)) (\cos (x) - \sin (x))) = 1$

Step 5.5.2

Remove unnecessary parentheses.

$x' (\cos (x) + \sin (x)) (\cos (x) - \sin (x)) = 1$

Step 5.6

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

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Step 5.6.1

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

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

Step 5.6.2

Simplify the left side.

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Step 5.6.2.1

Cancel the common factor of $\cos (x) + \sin (x)$ .

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Step 5.6.2.1.1

Cancel the common factor.

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

Step 5.6.2.1.2

Rewrite the expression.

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

Step 5.6.2.2

Cancel the common factor of $\cos (x) - \sin (x)$ .

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Step 5.6.2.2.1

Cancel the common factor.

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

Step 5.6.2.2.2

Divide $x'$ by $1$ .

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

Step 6

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

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