Math differentiation

posted by .

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 -

    #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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Algebra II

    I'm not quite sure how to set this problem up. This is the first of the entire worksheet and I'm sure if I understand one problem I can finish the rest pretty easily. The problem states: When there is no wind, Amelia flies her plane …
  2. Calculus

    Hi, I'm having trouble using implicit differentiation to determine dy/dx in the form dy/dx = f(x,y) for sin(2x+3y)=3x^3y^2+4 Do I make it sin(2x+3y)-3x^3y^2=4 then differentiate to get 2cos(2x+3y)-9x^2*2y=0 ?
  3. Public High School Pre Calculus HH

    Ok I do not know how to do this problem. I know that csc is simply sin^-1 sin (- pi/12) csc ( (37 pi)/12 ok now I know this also sin (- pi /12 ) = - sin ( pi/12) not really sure how that helps I don't know were to go from here ( - …
  4. Trigonometry

    How do you factor and simplify, 1-sin^3x and sec^4x-2sec^2xtan^2x+tan^4x
  5. trig

    prove: tan*2x-sin*2x=sin*2xtan*2x
  6. trig

    prove: tan*2x-sin*2x=sin*2xtan*2x
  7. Math-Derivatives.

    if y=sec^(3)2x, then dy/dx= my answer is 6 sec^3 2xtan 2x but im not sure if that is right. ?
  8. Calculus

    Find dy/dx for y=sin(x+y) A. 0 B. (cos(x+y))/(1-cos(x+y) C. cos(x+y) D. 1 E. None of these I know I'm supposed to use implicit differentiation but I'm not sure how to go about it with sin
  9. Math

    Let K(x)= (sin(3x-7)) / (e^x-2) (I'm sorry if it is not clear here, but the denominator in K(x) has the natural number, e, to the power of just x. So "x-2" is not the exponent. "x" is the exponent). The Question: Does K(x) have a horizontal …
  10. Math (Integrals) (Using given method)

    I would like to solve the ∫sin^2(pix) dx Using the given substitution identity: sin^2(x) = (1/2)(1-cos2x) This is what I did so far: ∫sin^2(pix) let u = pix du = pi dx (1/pi)∫sin^2(u)du Applying the identity is where I'm lost …

More Similar Questions