Homework Help: Math differentiation

Posted by Marissa on Wednesday, August 27, 2008 at 1:52am.

1. y = x^x
2. y = (sin^2xtan^4x)/(x^2+1)^2
3. y = x^(1/x)
4. y = 2^(-x^3)

Okay, so I'm pretty sure 1 is done easily:
ln y = ln (x^x)
ln y = x ln x
d/dx ln y = d/dx x ln x
y'(1/y) = x(1/x) + ln x(1) = 1 + ln x
y'(1/y) = ln x + 1
y' y(1/y) = (ln x + 1)y
y' = (ln x + 1)y

since y = x^x,
y' = (ln x + 1)x^x

For #2, this is as far as I got:
ln y = ln[(sin^2xtan^4x)/(x^2+1)^2
y'(1/y) = ln(sin^2xtan^4x) - ln(x^2+1)^2

What do I do after that?

#3, I thought was like #1, but I still think my answer is incorrect or maybe I didn't simplify properly:
ln y = (1/x) ln x
d/dx ln y = d/dx (1/x) ln x
y'(1/y) = (1/x)*(1/x) + ln x*(-1/x^2) = (1/x)^2 + ln x*(-1/x^2)

y' = (1/x)^2 + ln x*(-1/x^2)y

y' = (1/x)^2 + ln x*(-1/x^2)x^(1/x)

And for #4 I am completely lost since there is an exponent within the exponent. Any kind of hints or clues would be SO appreciated.

Math differentiation - Reiny, Wednesday, August 27, 2008 at 8:48am
#1 is good

in #2 you could go further before taking any derivative
y'(1/y) = ln(sin^2xtan^4x) - ln(x^2+1)^2 or
y'(1/y) = ln(sin^2x) + ln(tan^4x) - 2ln(x^2 + 1)
y'(1/y) = 2ln(sinx) + 4ln(tanx) - 2ln(x^2+1)

and now differentiate it

In #3 from your second last line of
y' = (1/x)^2 + ln x*(-1/x^2)y I would do

y ' = y(1 - lnx)/x^2 and leave it like that

#4 is exponential differentiation.
Notice the base is a constant, not a variable

recall if y = a^u
then y' = (ln a)(a^u)(u')

so for y = 2^(-x^3)
y' = ln2(-3x^2)(2^(-x^3))
= -3ln2(x^2)(2^(-x^3))

If you want to do it by the same method as the above, you would get

lny = ln(2^(-x^3))
= (-x^3)(ln2)
= -ln2(x^3)
now take the derivative

y'/y = -ln2(3)x^2
y' = -3ln2(x^2)y
= my answer above

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