Let y = lnx/x^5 find a)find y'(x). b)find y''(x)?

Answer 1

y' = (1 - 5lnx) / x^6

y" = (30lnx - 11) / x^7



Here's the solution.

y = lnx/x^5 can also be written as y=(lnx)(x^-5), right? The product rule will be much easier to use than the quotient rule in this case.

Hence, we get:

y = (lnx)(x^-5)

y' = (lnx)(-5x^-6) + (x^-5)(1 / x)

y' = (lnx)(-5 / x^6) + (1 / x^6)

y' = (-5lnx / x^6) + (1 / x^6)

.:. y' = (1 - 5lnx) / x^6



To find the second derivative, let us, again, rewrite the first derivative in such a way that we will be using the product rule instead. We get:

y' = (1 - 5lnx) (x^-6)

y" = (1 - 5lnx) (-6x^-7) + (x^-6) (-5 / x)

y" = [(1 - 5lnx) (-6) / x^7] - (5 / x^7)

y" = [-6(1 - 5lnx) - 5] / x^7

y" = (-6 +30lnx - 5) / x^7

.:. y" = (30lnx - 11) / x^7

Answer 2

lnx/x^5

y '= x^5(1/x) - 5x^4lnx
= x^4 - 5x^4lnx

y " = 4x^3 - 5(x^4)(1/x) - 5(lnx)(4x^3)
= 4x^3 - 5x^3 - 20x^3lnx
= -x^3 - 20x^3lnx
= -x^3(1 + 20lnx)

Answer 3

Use the quotient rule.

y = lnx/x^5

dy/dx = [(x^5)(1/x) - (ln x)(5x^4)] / (x^5)²

dy/dx = [x^4 - (5x^4)(ln x)] / x^10

dy/dx = [1 - 5(ln x)] / x^6

Use the quotient rule again to get d²y/dx².

Answer 4

1. Split by parts,

ln(x)(1/x^5)'+(ln(x))'(1/x^5)=
ln(x)(-5/x^6)+(1/x)((1/x^5)=
1/x^6-(5ln(x)/x^6)

2. Do it again:

(1/x^6)'-5((ln(x)(1/x^6)'+((ln(x))'(1/x^6)=

-6/x^7-5(-6ln(x)/x^7+1/x^7)=

-11/x^7+30ln(x)/x^7

Answer 5

y=lnx/x^5

y'=(x^5)(ln x)' - (lnx)(x^5)'
________________

(x^5)^2

y'=x^4-x^5lnx
________
x^10

y'=(1-xlnx)/x^6

apply your derivative rules again to get y''