1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differentiate this Eqn

  1. May 28, 2009 #1
    1: U=e^(-2t)

    2: Y=e^(4(x^3)-3x)

    3: T= x^2/3 * e^-3x

    q4: find the equation of the curve passing through the point (1,2) and having the slope at the point (x,y) given by dy/dx = 3/x

    q5: find dy/dx when y^2 + (y^3)*(e^y) = xy + ln(3x)

    q6: Find f If F'(x) = 5e^x and f(0) = 2



    my awnsers

    1: (1/-2)ln(t)

    2: (1/ (4*-3)) * ln(x^3 - x)

    3: -3(ln(1/ (x^2/3))) ???

    4: (3/2)x^2 * (ln(x)) + c ???

    5: this is implicit differentiation? not sure how to solve this one

    6: i don't know how to intergrate an exponential function
     
  2. jcsd
  3. May 28, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    Could you show us how you arrived at those answers? They are unfortunately all wrong. For question 6 you do not have to integrate.
     
  4. May 28, 2009 #3

    Mark44

    Staff: Mentor

    You are mistakenly thinking that d/dx(e^x) = ln x, which is not true. These functions are inverses of one another, but neither is the derivative of the other.

    The formula you want is this: d/dx(e^u) = e^u * du/dx, with u being a differentiable function of x.
     
  5. May 28, 2009 #4

    Cyosis

    User Avatar
    Homework Helper

    On second thought, regarding problem 6, do you mean F(0)=2?
     
  6. May 28, 2009 #5

    Mark44

    Staff: Mentor

    That has to be what Vorcil meant. As stated, "Find f If F'(x) = 5e^x and f(0) = 2," there is no solution. The problem should be stated as
    Find F if F'(x) = 5e^x and F(0) = 2​
    or as

    Find f if f'(x) = 5e^x and f(0) = 2​

    Either of these ways establishes that one of the functions is the derivative of the other. Without this information, you're stuck.
     
  7. May 28, 2009 #6
    1:u=e^-2t
    isn't it just e^-2t?
    oh chain rule, -2*(e^-2t)



    2: e^((4x^3)-3x)
    don't know, or am not sure how this is differentiated, do i use the chain rule?
    e^((4x^3)-3x) * ((12x^2) -3)

    3: y= x^2/3 * e^-3x
    not sure, do i use thr product rule and chain rule for this

    ((e^-3x)*(-3))-chain rule
    so
    (x^2/3) * ((e^-3x)*(-3)) + (2/3)x^(-1/3)*e^(-3x)

    those are the first three
     
    Last edited: May 28, 2009
  8. May 28, 2009 #7
    how do i figure out the equation of a curve if, 3/x = dy/dx, given the point 1,2
     
  9. May 28, 2009 #8

    Cyosis

    User Avatar
    Homework Helper

    First three are correct. You have y' and you want to know y. How do you obtain the original function save a constant if you're given the derivative?
     
  10. May 28, 2009 #9
    integrate?

    3/x = 3x*ln(x) +c?
     
  11. May 28, 2009 #10

    Cyosis

    User Avatar
    Homework Helper

    Yes, but you still need to determine c.
     
  12. May 28, 2009 #11
    3*1 * ln1 = 0
    do c = +2

    3x * ln(x) + 2 = y

    how do i do that other one, find f if f'(x) = 5e^x and f(0) = 2
     
  13. May 28, 2009 #12
    5e^x

    so

    5e^0 + c

    5e^0 = 5 so + c= -3

    5e^0 -3 = y
     
  14. May 28, 2009 #13

    Mark44

    Staff: Mentor

    This is wrong. You're trying to answer the question d/dx(???) = 3/x.

    Do you know any function whose derivative is 1/x? Now how about 3/x? It is absolutely not 3x*ln(x) with or without the added constant.
     
  15. May 28, 2009 #14

    Cyosis

    User Avatar
    Homework Helper

    The exact same way as the previous one.

    This is not right. Sorry for saying it was, but I overlooked it at first glance, where does the x come from? Secondly it is pretty horrible math to put an equal sign in between those two expressions. Write " the integral of 3/x=....' or use latex.
     
  16. May 28, 2009 #15
    oh so it's just 3*(ln(x)+c?
     
  17. May 28, 2009 #16

    Cyosis

    User Avatar
    Homework Helper

    While the answer is somewhat right you should really put some effort into noting down math. I hope you don't write it down like this on paper or on an exam, because it is going to cost you points.

    The correct way of doing it would be:

    [tex]
    \begin{align*}
    & f'(x)=5 e^x
    \\
    & f(x)=\int f'(x) dx = \int 5 e^x dx=5 e^x +c
    \\
    & f(0)=2=5+c \Rightarrow c=-3
    \\
    & f(x)=5e^x-3
    \end{align*}
    [/tex]
     
  18. May 28, 2009 #17

    Mark44

    Staff: Mentor

    Yes, and to make a complete thought, you can say this:

    d/dx(3*(ln(x)) = 3/x

    or

    [itex]\int 3/x * dx = 3 ln(x) + C[/itex]
     
  19. May 28, 2009 #18
    yeah i do do it properly in exams, i just don't know how to on computer
     
  20. May 28, 2009 #19

    Cyosis

    User Avatar
    Homework Helper

    [tex.]
    \begin{align*}
    & f'(x)=5 e^x
    \\
    & f(x)=\int f'(x) dx = \int 5 e^x dx=5 e^x +c
    \\
    & f(0)=2=5+c \Rightarrow c=-3
    \\
    & f(x)=5e^x-3
    \end{align*}
    [/tex.]

    Here is the code so you can take a look. The dots within the tex brackets should not be there if you want it to be showed correctly.
     
  21. May 28, 2009 #20

    Mark44

    Staff: Mentor

    The first form I did was just plain text, but I showed that I was taking the derivative on the thing on the left side.

    The second form used LaTeX to draw an integral symbol. It looks like this the way I typed it:
    [itex ]\int 3/x * dx = 3 ln(x) + C[ /itex]

    I put in an extra space after itex in the first tag, and another in the second tag so that the browser wouldn't render it. The itex and /itex pairs are for LaTeX that is inline with a sentence.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Differentiate this Eqn
Loading...