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How do you differentiate #f(x)= sqrt (ln(1/x)/x^2#?

Answer 1

Christopher Allers

Using chain and quotient rule.

Chain rule states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#

Quotient rule states that for #y=f(x)/g(x)#, #y'=(f'g-fg')/g^2#

We can go renaming parts of the expression in order to make up the chain.

#f(x)=sqrt(u)# #u=ln(v)/x^2# #v=1/x#

#(df(x))=(dx)=1/(2u^(1/2))*((1/v)(-1/x^2)(x^2)-ln(1/x)2x)/x^4#

Substituting #v#, then #u# and aggregating what's possible, so far

#(df(x))/(dx)=1/(2u^(1/2))*((1/(1/x))(-1/cancel(x^2))(cancel(x^2))-ln(1/x)2x)/x^4#

#(df(x))/(dx)=1/(2((ln(1/x))/x^2)^(1/2))*((x)(-1/cancel(x^2))(cancel(x^2))-ln(1/x)2x)/x^4#

#(df(x))/(dx)=(-x-2xln(1/x))/(2x^4sqrt((ln(1/x))/x^2))=color(green)((-1-2ln(1/x))/(2x^3sqrt((ln(1/x))/x^2)))#

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Answer 2

Ezra Adams

To differentiate f(x) = sqrt(ln(1/x)/x^2), you would use the chain rule and the power rule. The derivative of f(x) is given by:

f'(x) = (1/2) * (ln(1/x)/x^2)^(-1/2) * [(1/x^2) * ((-1/x) * (1/x) - ln(1/x) * (-1/x^2))] + sqrt(ln(1/x)/x^2) * [(-1/x^2) - 2 * ln(1/x)/x^3]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.