# a) Find y' given y= (x)sec(3x) b) Find y' given y= [(x)/(x+2)]^6

## Question:

a) Find {eq}\displaystyle {y}' {/eq} if {eq}y=x\sec (3x) {/eq}

b) Find {eq}\displaystyle {y}' {/eq} if {eq}y=\left (\frac{x}{x+2} \right )^{6} {/eq}

## Differentiation using the Product Rule and Quotient Rule of Differentiation

If we have two functions of {eq}x {/eq}, say {eq}u(x) {/eq} and {eq}v(x), {/eq} then the product rule of differentiation states that

$$\frac{d}{dx}\left(u(x)v(x)\right) = u(x) v'(x) + v(x)u'(x)$$

and the quotient rule of differentiation states that

$$\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{v(x)u'(x) - u(x)v'(x)}{(v'(x))^2}$$

We can solve this problem by using these rules.

## Answer and Explanation:

#### Part (a)

Given: {eq}y = x\sec (3x) {/eq}

We use the product rule to differentiate this function.

Let say that $$u = x, v = \sec (3x)\\$$

It implies that:

$$u' = 1 \\ v' = 3 \sec(3x)\tan(3x)$$

Thus, by substitution, we can get:

\begin{align*} y' &=1 \cdot \sec(3x) + x \cdot 3\cdot \sec(3x)\tan(3x) &\text{[Using the product rule and chain rule]} \\ &=\sec(3x) (1+ 3x\tan(3x)) \end{align*}

#### Part (b)

Given: {eq}\displaystyle {y}'\ if \ y=\left (\frac{x}{x+2} \right )^{6} {/eq}

Step 1. Apply Chain Rule to {eq}y = \left(\frac{x}{x+2}\right)^6 {/eq}

Note that: {eq}\frac{dx^n}{dx} = nx^{n-1} {/eq}

Then,

$$y' = 6\cdot (\frac{x}{x+2})^5\cdot \frac{d(\frac{x}{x+2})}{dx}$$

Step 2. To evaulate {eq}\frac{d\left(\frac{x}{x+2}\right)}{dx} {/eq} , we use quotient rule of differentiation with {eq}u = x, v = x+2. {/eq}

Thus,

\begin{align*} \frac{d(\frac{x}{x+2})}{dx} &= \frac{(x+2) \cdot 1 - x \cdot 1}{(x+2)^2} &\text{[Apply Quotient Rule]} \\ &=\frac{2}{(x+2)^2} \end{align*}

Step 3 Combine the two expressions.

Therefore,

\begin{align*} y' &= 6\cdot \left(\frac{x}{x+2}\right)^5\cdot \frac{2}{(x+2)^2}\\ &= \frac{6x^5}{(x+2)^7} \end{align*}