Asked by TayB
Use logarithmic differentiation to find the derivative of the following equation.
y = (2x+1)^5(x^4−3)^6
y = (2x+1)^5(x^4−3)^6
Answers
Answered by
Izzy
y=(2x+1)^5*(x^4-3)^6 {take the ln of both sides}
ln(y) = ln((2x+1)^5*(x^4-3)^6) {simplify the right side}
ln(y) = ln((2x+1)^5) + ln((x^4-3)^6) {simplify further}
ln(y) = 5ln(2x+1) + 6ln(x^4-3) {take the derivative} {lets do each term individually}
Deriviative of ln(y) =(1/y)y' {deriviative of ln(y) = 1/y}
Deriviative of 5ln(2x+1) = 5*2/(2x+1) = 10/(2x+1) {Use chain rule, let u=2x+1; du=2}
Deriviative of 6ln(x^4-3) = 6*4x^3/(x^4-3) = 24x^3/(x^4-3) {Use chain rule, let u=x^4-3; du=4x^3}
{now put it together}
(1/y)y' = 10/(2x+1) + 24x^3/(x^4-3) {multiply both sides by y}
y' = y * (10/(2x+1) + 24x^3/(x^4-3)) {sustitute y for (2x+1)^5*(x^4-3)^6 (the original equation)}
y' = (2x+1)^5*(x^4-3)^6 * (10/(2x+1) + 24x^3/(x^4-3)) {simplify}
y' = 10(2x+1)^4*(x^4-3)^6 + (24x^3)(2x+1)^5*(x^4-3)^5 {that's about as far as you go}
ln(y) = ln((2x+1)^5*(x^4-3)^6) {simplify the right side}
ln(y) = ln((2x+1)^5) + ln((x^4-3)^6) {simplify further}
ln(y) = 5ln(2x+1) + 6ln(x^4-3) {take the derivative} {lets do each term individually}
Deriviative of ln(y) =(1/y)y' {deriviative of ln(y) = 1/y}
Deriviative of 5ln(2x+1) = 5*2/(2x+1) = 10/(2x+1) {Use chain rule, let u=2x+1; du=2}
Deriviative of 6ln(x^4-3) = 6*4x^3/(x^4-3) = 24x^3/(x^4-3) {Use chain rule, let u=x^4-3; du=4x^3}
{now put it together}
(1/y)y' = 10/(2x+1) + 24x^3/(x^4-3) {multiply both sides by y}
y' = y * (10/(2x+1) + 24x^3/(x^4-3)) {sustitute y for (2x+1)^5*(x^4-3)^6 (the original equation)}
y' = (2x+1)^5*(x^4-3)^6 * (10/(2x+1) + 24x^3/(x^4-3)) {simplify}
y' = 10(2x+1)^4*(x^4-3)^6 + (24x^3)(2x+1)^5*(x^4-3)^5 {that's about as far as you go}
Answered by
Steve
you can factor out the 2(2x+1)^4 (x^4-3)^5
and you have
2(2x+1)^4 (x^4-3)^5 (5(x^4-3) + 12x^3(2x+1))
= 2(2x+1)^4 (x^4-3)^5 (29x^4+12x^3-15)
and you have
2(2x+1)^4 (x^4-3)^5 (5(x^4-3) + 12x^3(2x+1))
= 2(2x+1)^4 (x^4-3)^5 (29x^4+12x^3-15)
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