Question

find y' and y''.
y=(ln x)/(x^2)

Answer 1

if \[y = \frac {\ln x}{x^2}\] u can also write it as \[y = \ln x \times x^{-2}\] if it makes it easier for u to think about it?

so are u familiar with the product rule? @muneka.fea2956 ?

if u call ln x --> "u" and x^-2 --> "v"

product rule states:
\[y = v \ \times u \rightarrow so \rightarrow y' = (u' \times v) + (v' \times u)\]

is this makin sense so far?
sorry, it takes me a while to type in latex

Answer 2

so back to the above:
v = x^-2
u = ln x
v' = -2x^-3
u' = 1/x

so add them all together as per the product rule
y′=(u′×v)+(v′×u)
= (1/x) times (1/x^2) + ( -2/x^3) times (ln x)
= 1/x^3 + -2 ln x / x^3
= 1 - 2 ln x
--------------
x^3
or eqn form: \[y' = \frac {1 - 2 \ln x}{x^3} \]

Answer 3

so y'' ... just use product rule again to find derivative of above @muneka.fea2956

Answer 4

this time u = (1 - 2 ln x)
and v = x^-3

Answer 5

thank you!! :)

Answer 6

all good ;D