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Mathematics 17 Online
OpenStudy (anonymous):

find the derivative: 3lnx/x^2

OpenStudy (anonymous):

applying quotient rule as follows

OpenStudy (anonymous):

\[d/dx(3lnx/x^2)\]

OpenStudy (anonymous):

{[3x]-6xlnx}/x^{4}

OpenStudy (anonymous):

hmmmm I don't know if that's right nor not :)

OpenStudy (anonymous):

d1st(2nd) -d2nd(1st)/2nd^2 is the formula right?

OpenStudy (anonymous):

yes but denominator has whole square of function in denominator

OpenStudy (anonymous):

i am writing in detail just a sec

OpenStudy (anonymous):

d/d((f(x))/g(x) =(f^' (x)g(x)-f(x) g^' (x))/〖{g(x)}〗^2

OpenStudy (anonymous):

OpenStudy (anonymous):

what is meant by the word differentiable and to differentiate?

OpenStudy (anonymous):

[3x-6xlnx]/x^4 taking 3x common and cancelling x finally 3[1-6lnx]/x^3

OpenStudy (anonymous):

differentiate means finding derivative simply by applying formulae already known and differentiable function means a function who's right derivative and left derivative exist

OpenStudy (anonymous):

In calculus (a branch of mathematics), a differentiable function is a function whose derivative exists at each point in its domain.

OpenStudy (anonymous):

a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's instantaneous velocity.

OpenStudy (anonymous):

ok but what I don't follow here is two things 1) how you handled the numerator and 2) how you ended up with x^3 in the denominator

OpenStudy (anonymous):

let us denote numerator by f(x) and denominator by g(x) then in numerator we have g(x)f'(x)-f(x)g'(x) where f'(x) and g'(x) means their derivatives in denominator we have [g(x)]^2 right this is the quotient rule, if u get it remaining is too easy

OpenStudy (anonymous):

i am finding difficulty in typing fractions here

OpenStudy (anonymous):

I get it but I still don't get how to find the derivative of 3lnx

OpenStudy (anonymous):

f(x)=3lnx so f'(x)=3/x because derivative of lnx is 1/x

OpenStudy (anonymous):

the quotient rule I get but I don't get how to find the derivative of a natural log or a regular log for that matter

OpenStudy (anonymous):

and g(x)=x^2 so derivative g'(x)=2x by power rule

OpenStudy (anonymous):

putting the values and simplifying u get the answer

OpenStudy (anonymous):

ok I think I remember lnx=1/x, so 3lnx=3/x?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

I still don't see how you got 3(1-6lnx) in the numerator if the derivative of the numerator is 3/x

OpenStudy (anonymous):

in numerator putting values we get numerator as 3(x^2)x-(3lnx)(2x) =3x-6xlnx =3x(1-2lnx) sorry i wrote 6 earlier:D then cancel this x of numerator with one of x in denominator

OpenStudy (anonymous):

ok I worked it on paper and I got.....|dw:1319795677700:dw|

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