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Differentiate this Eqn

  • Thread starter vorcil
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  • #1
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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
 

Answers and Replies

  • #2
Cyosis
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Could you show us how you arrived at those answers? They are unfortunately all wrong. For question 6 you do not have to integrate.
 
  • #3
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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.
 
  • #4
Cyosis
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On second thought, regarding problem 6, do you mean F(0)=2?
 
  • #5
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On second thought, regarding problem 6, do you mean F(0)=2?
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.
 
  • #6
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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:
  • #7
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how do i figure out the equation of a curve if, 3/x = dy/dx, given the point 1,2
 
  • #8
Cyosis
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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?
 
  • #9
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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?
integrate?

3/x = 3x*ln(x) +c?
 
  • #10
Cyosis
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Yes, but you still need to determine c.
 
  • #11
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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
 
  • #12
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5e^x

so

5e^0 + c

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

5e^0 -3 = y
 
  • #13
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integrate?

3/x = 3x*ln(x) +c?
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.
 
  • #14
Cyosis
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The exact same way as the previous one.

vorcil said:
3/x = 3x*ln(x) +c?
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.
 
  • #15
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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.
oh so it's just 3*(ln(x)+c?
 
  • #16
Cyosis
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vorcil said:
5e^x

so

5e^0 + c

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

5e^0 -3 = y
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]
 
  • #17
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oh so it's just 3*(ln(x)+c?
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]
 
  • #18
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yeah i do do it properly in exams, i just don't know how to on computer
 
  • #19
Cyosis
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[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.
 
  • #20
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yeah i do do it properly in exams, i just don't know how to on computer
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.
 
  • #21
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thank you =] i don't mean to be so questionative or demmanding when i ask questions
 
  • #22
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No problem. And a thank you goes a long way. I think you have to be "questionative" when you ask questions. Pretty sure that's not a word, though.:smile:
 

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