Find the derivative of the function
P(x)=ln [ (4x + 1)^3 / (2x − 5)^4 ]
is
a. −4(2x−17) / (4x+1)(2x−5)
b.−4(2x−17) / (4x+1)
c. 4(−2x−17) / (4x+1)(2x−5)
d. (−2x−17) / (4x+1)(2x−5)
P’(x)=(ln[(4x+1)3/(2x−5)4])’=1(4x+1)3/(2x−5)4∗((4x+1)3/(2x−5)4)’=(2x−5)4(4x+1)3∗12(4x+1)2∗(2x−5)4−8(2x−5)3∗(4x+1)3(2x−5)8=12(2x−5)−8(4x+1)(4x+1)(2x−5)=24x−60−32x−8(4x+1)(2x−5)=4(−2x−17)(4x+1)(2x−5)P’(x)=(ln [ (4x + 1)^3 / (2x − 5)^4 ])’= \frac{1}{(4x + 1)^3 / (2x − 5)^4}*((4x + 1)^3 / (2x − 5)^4)’=\frac{(2x-5)^4}{(4x+1)^3}*\frac{12(4x+1)^2*(2x-5)^4-8(2x-5)^3*(4x+1)^3}{(2x-5)^8}=\frac{12(2x-5)-8(4x+1)}{(4x+1)(2x-5)}=\frac{24x-60-32x-8}{(4x+1)(2x-5)}=\frac{4(-2x-17)}{(4x+1)(2x-5)}P’(x)=(ln[(4x+1)3/(2x−5)4])’=(4x+1)3/(2x−5)41∗((4x+1)3/(2x−5)4)’=(4x+1)3(2x−5)4∗(2x−5)812(4x+1)2∗(2x−5)4−8(2x−5)3∗(4x+1)3=(4x+1)(2x−5)12(2x−5)−8(4x+1)=(4x+1)(2x−5)24x−60−32x−8=(4x+1)(2x−5)4(−2x−17)
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