Find the Derivative - d/dx y=(sin(2x))/(1+cos(2x))

$y = \frac{\sin (2 x)}{1 + \cos (2 x)}$

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = \sin (2 x)$ and $g (x) = 1 + \cos (2 x)$ .

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

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

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

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

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

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

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

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

Differentiate.

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

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

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

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

Simplify the expression.

Multiply $2$ by $1$ .

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

Move $2$ to the left of $(1 + \cos (2 x)) \cos (2 x)$ .

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

By the Sum Rule, the derivative of $1 + \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\cos (2 x)]$ .

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

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

$\frac{2 (1 + \cos (2 x)) \cos (2 x) - \sin (2 x) (0 + \frac{d}{d x} [\cos (2 x)])}{{(1 + \cos (2 x))}^{2}}$

Add $0$ and $\frac{d}{d x} [\cos (2 x)]$ .

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

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

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

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

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

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

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

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

Multiply.

Multiply $- 1$ by $- 1$ .

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

Multiply $\sin (2 x)$ by $1$ .

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

Raise $\sin (2 x)$ to the power of $1$ .

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

Raise $\sin (2 x)$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

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

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

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

$\frac{2 (1 + \cos (2 x)) \cos (2 x) + \sin^{2} (2 x) (2 \cdot 1)}{{(1 + \cos (2 x))}^{2}}$

Simplify the expression.

Multiply $2$ by $1$ .

$\frac{2 (1 + \cos (2 x)) \cos (2 x) + \sin^{2} (2 x) \cdot 2}{{(1 + \cos (2 x))}^{2}}$

Move $2$ to the left of $\sin^{2} (2 x)$ .

$\frac{2 (1 + \cos (2 x)) \cos (2 x) + 2 \cdot \sin^{2} (2 x)}{{(1 + \cos (2 x))}^{2}}$

Simplify.

Apply the distributive property.

$\frac{(2 \cdot 1 + 2 \cos (2 x)) \cos (2 x) + 2 \sin^{2} (2 x)}{{(1 + \cos (2 x))}^{2}}$

Apply the distributive property.

$\frac{2 \cdot 1 \cos (2 x) + 2 \cos (2 x) \cos (2 x) + 2 \sin^{2} (2 x)}{{(1 + \cos (2 x))}^{2}}$

Simplify the numerator.

Simplify each term.

Multiply $2$ by $1$ .

$\frac{2 \cos (2 x) + 2 \cos (2 x) \cos (2 x) + 2 \sin^{2} (2 x)}{{(1 + \cos (2 x))}^{2}}$

Multiply $2 \cos (2 x) \cos (2 x)$ .

Raise $\cos (2 x)$ to the power of $1$ .

$\frac{2 \cos (2 x) + 2 (\cos^{1} (2 x) \cos (2 x)) + 2 \sin^{2} (2 x)}{{(1 + \cos (2 x))}^{2}}$

Raise $\cos (2 x)$ to the power of $1$ .

$\frac{2 \cos (2 x) + 2 (\cos^{1} (2 x) \cos^{1} (2 x)) + 2 \sin^{2} (2 x)}{{(1 + \cos (2 x))}^{2}}$

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

$\frac{2 \cos (2 x) + 2 {\cos (2 x)}^{1 + 1} + 2 \sin^{2} (2 x)}{{(1 + \cos (2 x))}^{2}}$

Add $1$ and $1$ .

$\frac{2 \cos (2 x) + 2 \cos^{2} (2 x) + 2 \sin^{2} (2 x)}{{(1 + \cos (2 x))}^{2}}$

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

$\frac{2 \cos (2 x) + 2 (\cos^{2} (2 x)) + 2 \sin^{2} (2 x)}{{(1 + \cos (2 x))}^{2}}$

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

$\frac{2 \cos (2 x) + 2 (\cos^{2} (2 x)) + 2 (\sin^{2} (2 x))}{{(1 + \cos (2 x))}^{2}}$

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

$\frac{2 \cos (2 x) + 2 (\cos^{2} (2 x) + \sin^{2} (2 x))}{{(1 + \cos (2 x))}^{2}}$

Rearrange terms.

$\frac{2 \cos (2 x) + 2 (\sin^{2} (2 x) + \cos^{2} (2 x))}{{(1 + \cos (2 x))}^{2}}$

Apply pythagorean identity.

$\frac{2 \cos (2 x) + 2 \cdot 1}{{(1 + \cos (2 x))}^{2}}$

Multiply $2$ by $1$ .

$\frac{2 \cos (2 x) + 2}{{(1 + \cos (2 x))}^{2}}$

Reorder terms.

$\frac{2 + 2 \cos (2 x)}{{(1 + \cos (2 x))}^{2}}$

Factor $2$ out of $2 + 2 \cos (2 x)$ .

Factor $2$ out of $2$ .

$\frac{2 \cdot 1 + 2 \cos (2 x)}{{(1 + \cos (2 x))}^{2}}$

Factor $2$ out of $2 \cdot 1 + 2 \cos (2 x)$ .

$\frac{2 (1 + \cos (2 x))}{{(1 + \cos (2 x))}^{2}}$

Cancel the common factor of $1 + \cos (2 x)$ and ${(1 + \cos (2 x))}^{2}$ .

Factor $1 + \cos (2 x)$ out of $2 (1 + \cos (2 x))$ .

$\frac{(1 + \cos (2 x)) \cdot 2}{{(1 + \cos (2 x))}^{2}}$

Cancel the common factors.

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

$\frac{(1 + \cos (2 x)) \cdot 2}{(1 + \cos (2 x)) (1 + \cos (2 x))}$

Cancel the common factor.

$\frac{(1 + \cos (2 x)) \cdot 2}{(1 + \cos (2 x)) (1 + \cos (2 x))}$

Rewrite the expression.

$\frac{2}{1 + \cos (2 x)}$

Find the Derivative - d/dx y=(sin(2x))/(1+cos(2x))

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