what is the first and second derivative for (lnx)/x^2 ?

Answer 1

y = (lnx)/x^2
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y = f(x)/g(x)
dy/dx = (f' *g - g' *f)/g^2
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dy/dx = [(1/x) *x^2 - (2x) lnx] / x^4
dy/dx = [x - (2x) lnx] / x^4
dy/dx = [1 - 2 lnx] / x^3 >>>>
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d^y/dx^2 = [-(2/x)*x^3 - 3x^2(1 - 2 lnx)] / x^6
d^y/dx^2 = - [2 + 3(1 - 2 lnx)] / x^4
d^y/dx^2 = - [5 -6 lnx)] / x^4
d^y/dx^2 = [- 5 +6 lnx] / x^4 >>>

>>> answers

Answer 2

Hello

First derivative;

[(1/x)*(1/x^2) - 2x*lnx]/x^4 = 1/x^7 - 2lnx/x^3

Will work on the second one!

Hope This Helps!!

Answer 3

Let f(x) = ln(x) / x^2

To differentiate this, you must use the quotient rule.

f'(x) = uv' - vu' / v^2
Let the numerator of the f(x) expression be u, and the denominator v.

thus;
u = ln x
u' = 1/x
v = x^2
v' = 2x

Substituting in the above expression for f'(x)

f'(x) = ln x*2x - x^2*(1/x) / (x^2)^2
= 2x*ln(x) -x / x^4

The second derivative of this would require another application of the quotient rule.

Answer 4

1/(x^3)-2*ln(x)/x^3

-5/(x^4)+6*ln(x)/x^4