Use logarithmic differentiation to find the derivative of the function. y=(2x+1)^5(x^4-8)^6. y'=?

Answer 1

Y = (2x+1)^5 * (x^4 - 8)^6

lnY = ln[ (2x+1)^5 * (x^4 - 8)^6 ]

lnY = 5 ln(2x+1) + 6 ln(x^4 - 8)

d/dx [lnY] = d/dx [ 5 ln(2x+1) + 6 ln(x^4 - 8) ]

(1/Y)Y' = 5 * 1/ (2x+1)*[2] + 6 * 1/(x^4 - 8) * [4x]

(1/Y)Y' = 10 / (2x+1) + 24x / (x^4 - 8)

Y' = Y * [ 10 / (2x+1) + 24x / (x^4 - 8) ]

Y' = [ (2x+1)^5 * (x^4 - 8)^6 ] * [ 10 / (2x+1) + 24x / (x^4 - 8) ]

Y' = 10*(2x+1)^4 * (x^4 - 8)^6 + 24x*(2x+1)^5 * (x^4 - 8)^5

Answer 2

lny = 5ln(2x + 1)+ 6ln(x⁴ - 8)

y'/y = 5/(2x + 1)·2 + 6/(x⁴ - 8)·4x³ = 10/(2x + 1) + 24x³/(x⁴ - 8);

now y' = [10/(2x + 1) + 24x³/(x⁴ - 8)](2x + 1)⁵(x⁴ - 8)⁶