Please help ?

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3 Answers
Dec 9, 2017

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 :)

Dec 9, 2017

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

Dec 9, 2017

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.