Wednesday

July 30, 2014

July 30, 2014

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

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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