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.