How do you calculate the derivate for #f(x)=[(x+2)/(x+1)](2x-7)#?

Question

Evelyn Agard · Accepted Answer

Using a combination of Product Rule and Quotient Rule.

Product Rule: If #f(x)=g(x)\cdot h(x)#, then #f^\prime(x)=g(x)\cdot h\prime (x)+h(x)\cdot g\prime (x)# Quotient Rule: If #f(x)=\frac{g(x)}{h(x)}# then #f^\prime(x)=\frac{h(x)\cdot g\prime (x)-g(x)\cdot h\prime (x)}{h(x)^2}# Now, you can express #f(x)# as #f(x)=\underbrace{g(x)\cdot h(x)}_ ext{product rule}#, where #g(x)=\underbrace{\frac{x+2}{x+1}}_ ext{quotient rule}# and #h(x)=2x-7# OR as #f(x)=\underbrace{\frac{g(x)}{h(x)}}_ ext{quotient rule}#, where #g(x)=\underbrace{(x+2)(2x-7)}_ ext{product rule}# and #h(x)=x+1#. Try Prepwell Calculus for Android.

Luke Alvarez · Answer

undefined To calculate the derivative of $ f(x) = \frac{{(x+2)}}{{(x+1)}}(2x-7) $, you can use the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Applying the product rule to $ f(x) $, we get:

$ f'(x) = \left(\frac{{d}}{{dx}}\left(\frac{{x+2}}{{x+1}}\right)\right)(2x-7) + \left(\frac{{x+2}}{{x+1}}\right)\left(\frac{{d}}{{dx}}(2x-7)\right) $

Now, differentiate each part separately:

$ \frac{{d}}{{dx}}\left(\frac{{x+2}}{{x+1}}\right) = \frac{{(x+1)(1) - (x+2)(1)}}{{(x+1)^2}} $

$ \frac{{d}}{{dx}}(2x-7) = 2 $

Plug these derivatives back into the equation:

$ f'(x) = \frac{{(x+1) - (x+2)}}{{(x+1)^2}}(2x-7) + \frac{{x+2}}{{x+1}}(2) $

Now, simplify:

$ f'(x) = \frac{{-1}}{{(x+1)^2}}(2x-7) + 2\frac{{x+2}}{{x+1}} $

$ f'(x) = \frac{{-2x+7}}{{(x+1)^2}} + 2\frac{{x+2}}{{x+1}} $

$ f'(x) = \frac{{-2x+7}}{{(x+1)^2}} + \frac{{2(x+2)}}{{x+1}} $

$ f'(x) = \frac{{-2x+7+2(x+2)}}{{(x+1)^2}} $

$ f'(x) = \frac{{-2x+7+2x+4}}{{(x+1)^2}} $

$ f'(x) = \frac{{11}}{{(x+1)^2}} $