SOLUTION: Let f(x)= (4x-5)^4/(7x+5)^5. Find f'(x)

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Question 1160315: Let f(x)= (4x-5)^4/(7x+5)^5. Find f'(x)
Found 2 solutions by MathLover1, solver91311:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Let+f%28x%29=+%284x-5%29%5E4%2F%287x%2B5%29%5E5
Find f'%28x%29

to do so, apply the Quotient Rule : %28+f%2Fg+%29'=(f'*g-g'* f)/g%5E2+



f'%28x%29=%28d%2Fdx%29%28%284+x+-+5%29%5E4%2F%287+x+%2B+5%29%5E5%29
f'%28x%29=
%28d%2Fdx%29%284+x+-+5%29%5E4=16%284+x+-+5%29%5E3
%28d%2Fdx%29%287+x+%2B+5%29%5E5=35%287+x+%2B+5%29%5E4
f'%28x%29 =
f'%28x%29 =.....simplify
f'%28x%29 =
f'%28x%29 =%2816%284+x+-+5%29%5E3%2A%287x+%2B+5%29-+35%2A%284+x+-+5%29%5E4%29%2F%287+x+%2B+5%29%5E6

f'%28x%29=%28%284+x+-+5%29%5E3%2816%287+x+%2B+5%29-+35%2A%284+x+-+5%29%29%29%2F%287+x+%2B+5%29%5E6

f'%28x%29 =%28%284+x+-+5%29%5E3%28112x+%2B+80-+140+x+%2B+175%29%29%2F%287+x+%2B+5%29%5E6

f'%28x%29 = %28%284+x+-+5%29%5E3+%28-28x+%2B255%29%29%2F%287+x+%2B+5%29%5E6......factor out negative sign in front of 28x

f'%28x%29= -%28%284+x+-+5%29%5E3+%2828x+-255%29%29%2F%287+x+%2B+5%29%5E6


Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Use the quotient rule:




John

My calculator said it, I believe it, that settles it