Expand each function (using the appropriate technique/formula) Compute

1) ${\displaystyle f\left(x\right)=\sin \left(2x\right)}$

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

To apply the Chain Rule, set ${\displaystyle u}$ as ${\displaystyle 2x}$.

${\displaystyle \frac{d}{du}\left[\sin \left(u\right)\right]\frac{d}{dx}\left[2x\right]}$

The derivative of ${\displaystyle \sin \left(u\right)}$ with respect to ${\displaystyle u}$ is ${\displaystyle \cos \left(u\right)}$.

${\displaystyle \cos \left(u\right)\frac{d}{dx}\left[2x\right]}$

Replace all occurrences of ${\displaystyle u}$ with ${\displaystyle 2x}$.

${\displaystyle \cos \left(2x\right)\frac{d}{dx}\left[2x\right]}$

Differentiate.

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

${\displaystyle \cos \left(2x\right)\left(2\frac{d}{dx}\left[x\right]\right)}$

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

${\displaystyle \cos \left(2x\right)(2\cdot 1)}$

Simplify the expression.

Multiply ${\displaystyle 2}$ by ${\displaystyle 1}$.

${\displaystyle \cos \left(2x\right)\cdot 2}$

Move ${\displaystyle 2}$ to the left of ${\displaystyle \cos \left(2x\right)}$.

$2\cos \left(2x\right)$

2) ${\displaystyle f\left(x\right)={(3x-2)}^3}$

Differentiate using the chain rule, which states that ${\displaystyle \frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]}$ is ${\displaystyle f\prime \left(g\left(x\right)\right)g\prime \left(x\right)}$ where f(x)=x3${\displaystyle f\left(x\right)=x^3}$ and ${\displaystyle g\left(x\right)=3x-2}$.

To apply the Chain Rule, set ${\displaystyle u}$ as ${\displaystyle 3x-2}$.

${\displaystyle \frac{d}{du}\left[u^3\right]\frac{d}{dx}[3x-2]}$

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{du}\left[u^n\right]}$ is ${\displaystyle n{u}^{n-1}}$ where ${\displaystyle n=3}$.

${\displaystyle 3u^2\frac{d}{dx}[3x-2]}$

Replace all occurrences of ${\displaystyle u}$ with ${\displaystyle 3x-2}$.

${\displaystyle 3{(3x-2)}^2\frac{d}{dx}[3x-2]}$

Differentiate.

By the Sum Rule, the derivative of ${\displaystyle 3x-2}$ with respect to ${\displaystyle x}$ is ${\displaystyle \frac{d}{dx}\left[3x\right]+\frac{d}{dx}[-2]}$.

${\displaystyle 3{(3x-2)}^2(\frac{d}{dx}\left[3x\right]+\frac{d}{dx}[-2])}$

Since ${\displaystyle 3}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle 3x}$ with respect to ${\displaystyle x}$ is ${\displaystyle 3\frac{d}{dx}\left[x\right]}$.

${\displaystyle 3{(3x-2)}^2(3\frac{d}{dx}\left[x\right]+\frac{d}{dx}[-2])}$

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

${\displaystyle 3{(3x-2)}^2(3\cdot 1+\frac{d}{dx}[-2])}$

Multiply ${\displaystyle 3}$ by ${\displaystyle 1}$.

${\displaystyle 3{(3x-2)}^2(3+\frac{d}{dx}[-2])}$

Since ${\displaystyle -2}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle -2}$ with respect to ${\displaystyle x}$ is ${\displaystyle 0}$.

${\displaystyle 3{(3x-2)}^2(3+0)}$

Simplify the expression.

$9(3x-2{)}^2$

3) ${\displaystyle f\left(x\right)={(x^2+2x+3)}^2}$

Differentiate using the chain rule, which states that ${\displaystyle \frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]}$ is ${\displaystyle f\prime \left(g\left(x\right)\right)g\prime \left(x\right)}$ where ${\displaystyle f\left(x\right)=x^2}$ and ${\displaystyle g\left(x\right)=x^2+2x+3}$.

To apply the Chain Rule, set ${\displaystyle u}$ as ${\displaystyle x^2+2x+3}$.

${\displaystyle \frac{d}{du}\left[u^2\right]\frac{d}{dx}[x^2+2x+3]}$

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

${\displaystyle 2u\frac{d}{dx}[x^2+2x+3]}$

Replace all occurrences of ${\displaystyle u}$ with ${\displaystyle x^2+2x+3}$.

${\displaystyle 2(x^2+2x+3)\frac{d}{dx}[x^2+2x+3]}$

Differentiate.

By the Sum Rule, the derivative of ${\displaystyle x^2+2x+3}$ with respect to ${\displaystyle x}$ is ${\displaystyle \frac{d}{dx}\left[x^2\right]+\frac{d}{dx}\left[2x\right]+\frac{d}{dx}\left[3\right]}$.

${\displaystyle 2(x^2+2x+3)(\frac{d}{dx}\left[x^2\right]+\frac{d}{dx}\left[2x\right]+\frac{d}{dx}\left[3\right])}$

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

${\displaystyle 2(x^2+2x+3)(2x+\frac{d}{dx}\left[2x\right]+\frac{d}{dx}\left[3\right])}$

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

${\displaystyle 2(x^2+2x+3)(2x+2\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[3\right])}$

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

${\displaystyle 2(x^2+2x+3)(2x+2\cdot 1+\frac{d}{dx}\left[3\right])}$

Multiply ${\displaystyle 2}$ by ${\displaystyle 1}$.

${\displaystyle 2(x^2+2x+3)(2x+2+\frac{d}{dx}\left[3\right])}$

Since ${\displaystyle 3}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle 3}$ with respect to ${\displaystyle x}$ is ${\displaystyle 0}$.

${\displaystyle 2(x^2+2x+3)(2x+2+0)}$

Add ${\displaystyle 2x+2}$ and ${\displaystyle 0}$.

${\displaystyle 2(x^2+2x+3)(2x+2)}$

Simplify.

Apply the distributive property.

${\displaystyle (2x^2+2\left(2x\right)+2\cdot 3)(2x+2)}$

Combine terms.

${\displaystyle (2x^2+4x+6)(2x+2)}$

Reorder the factors of ${\displaystyle (2x^2+4x+6)(2x+2)}$.

${\displaystyle (2x+2)(2x^2+4x+6)}$

Expand ${\displaystyle (2x+2)(2x^2+4x+6)}$ by multiplying each term in the first expression by each term in the second expression.

${\displaystyle 2x\left(2x^2\right)+2x\left(4x\right)+2x\cdot 6+2\left(2x^2\right)+2\left(4x\right)+2\cdot 6}$

Simplify each term.

${\displaystyle 4x^3+8x^2+12x+4x^2+8x+12}$

Add ${\displaystyle 8x^2}$ and ${\displaystyle 4x^2}$.

${\displaystyle 4x^3+12x^2+12x+8x+12}$

Add ${\displaystyle 12x}$ and ${\displaystyle 8x}$.

$4x^3+12x^2+20x+12$