Verified answer

first you need to know ln'(y) = 1/y * y' and ln (x) + ln(y) = ln (x*y), ln(x)^a = a * ln(x)

y = (2x + 1)^5(x^4 − 1)^6

=> ln (y) = ln [(2x + 1)^5(x^4 − 1)^6]

=> ln (y) = ln [(2x + 1)^5] + ln [(x^4 − 1)^6]

=> ln (y) = 5 * ln (2x + 1) + 6 * ln (x^4 − 1)

=> (1 / y) * y' = [5 * 2 / (2x + 1)] + [6 * 4x^3 / (x^4 − 1)]

since y = (2x + 1)^5(x^4 − 1)^6 ; you time y to each term,

y' = [5 * 2 / (2x + 1)] * y + [6 * 4x^3 / (x^4 − 1)] * y

y' = [5 * 2 / (2x + 1)] * [(2x + 1)^5(x^4 − 1)^6] + [6 * 4x^3 / (x^4 − 1)] * [(2x + 1)^5(x^4 − 1)^6]

y' = 10 * (2x + 1)^4 * (x^4 − 1)^6 + 24x^3 * (2x + 1)^5 * (x^4 − 1)^5

Hope this is clear!!