Find the derivatives

can you show me some steps so I can follow along? And please don't just list the rules or answer. Thanks.

y=(lnx)^2




y= xlnx-x





y=ln(lnx)



y= ln(1/x)

Question # 1


y=(lnx)^2


or written like this

y = ln(x) * ln(x)

Now use the product rule

y' = dy/dx[ ln(x) * ln(x)]

= ln(x) * 1/x + 1/x * ln(x)

= ln(x) / x + ln(x) / x

= 2 ln(x) / x

So the derivative, y' is

Answer: y' = [2 ln(x)] / x

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Question # 2

y= xlnx-x


y' = dy/dx[x*ln(x)] - dy/dx[x]

= [x * 1/x + ln(x) * 1 ] - 1

= [1 + ln(x)] - 1

= ln(x)

Answer: The derivative y' is y' = ln(x).
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Question # 3

y=ln(lnx)


To find the derivative here, we will use the chain rule since we have a composition of functions

Let f(x) = ln(x) and g(x) = ln(x)

Then y = f(g(x)) [since f(g(x)) = ln(ln(x)]

y' = dy/dx [ f(g(x))] where f(x) = ln(x) and g(x) = ln(x)

= f '(g(x)) * g'(x) where f'(x) = 1/x ; g(x) = ln(x) and g'(x) = 1/x

= 1/[ln(x)] * 1/x

= 1/[x * ln(x)]

Answer: So y' = 1/[x * ln(x)]

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Question # 4

y= ln(1/x)

Here we can use the property of logarithms to simplify this

Remember that

log(a/b) = log(a) - log(b)

so

y = ln(1/x) = ln(1) - ln(x)

since ln(1) = 0, we simply have

y = - ln(x)

and so the derivative, y' is

Answer: y' = -1/x

Method # 2: We can also solve this using the chain rule

y = ln(1/x)

Let f(x) = ln(x) and g(x) = 1/x

Then

y = f(g(x))

y' = f ' g(x) * g'(x)

= 1/(1/x) * -x^-2

= x * - (1/x^2)

= - 1/x

so

Answer: y' = -1/x

[same answer as we got above]