Question

Please help ?

Answer 1

See the following.

Explanation:

We can use quotient rule for this question.

`f(x)=g(x)/(h(x))`

`f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2`

In this question `f(u)=(2u+1)/(u-2)`,
we can let `g(u) =2u+1` and `h(u)=u-2` .

`f'(u)=((u-2)*(2u+1 )'-(2u+1)*(u-2)')/(u-2)^2`

`=((u-2)(2)-(2u+1)(1))/(u-2)^2`

`=(2u-4-2u-1)/(u-2)^2`

`=-5/(u-2)^2`

So, the answer is B.
Hope this can help you :)

Answer 2

See below.

Explanation:

Product Rule:

`f'(a*b)=b*f'(a)+a*f'(b)`

Rewriting:

`(2u+1)/(u-2)` as `(2u+1)(u-2)^(-1)`

`a=(2u+1)`

`b=(u-2)`

`d/dx(2u+1)=2`

`d/dx(u-2)^(-1)=-(u-2)^(-2)*1`

Using product rule:

`(u-2)^(-1)* 2+(2u+1)* -1(u-2)^(-2)`

`->=2/((u-2))-(2u+1)/(u-2)^2=(2(u-2)-(2u+1))/((u-2)^2)`

`->=(2u-4-2u-1)/(u-2)^2=-(5)/(u-2)^2`

Answer 3

B

`color(blue)(f'(u) = -5/((u-2)^2)`

Explanation:

We are given a rational function `color(green)(y=f(u) = (2u+1)/(u-2))` `color(red)(Expression.1)`

We need to find the First Derivative of `f(u)`

By observing `f(u)` we know that we must use the Quotient Rule to differentiate .

Quotient Rule for finding the derivatives states that

`color(blue)((dy)/(du)[f(u)/g(u)] = [(g(u)*f'(u) - f(u)*g'(u))/[g(u)]^2]`

Using the Quotient Rule, we can write our `color(red)(Expression.1)` as

`(d/(du)(2u+1)(u-2) - (2u+1)(d/(du))(u-2)]/(u-2)^2` `color(red)(Expression.2)`

We can simplify and rewrite `color(red)(Expression.2)` as

`rArr`

`[(2d/(du)(u)+d/(du)(1))(u-2)-(d/(du)(u)+d/(du)(-2))(2u+1)]/(u-2)^2`

When we differentiate a constant we get zero(0).

`rArr[ (2*1+0)(u-2)-(1+0)(2u+1)]/(u-2)^2`

We can simplify this expression further as

`[2*(u - 2)-1*(2u+1)]/(u-2)^2`

`rArr(2u - 4-2u-1)/(u-2)^2`

We can simplify as

`rArr(cancel(2u) - 4-cancel(2u)-1)/(u-2)^2`

`rArr -(5)/(u-2)^2`

Hence,

`color(blue)(f'(u) = -5/((u-2)^2)`

I hope you find this solution process helpful.