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- #1

#### coolbeans33

##### New member

- Sep 19, 2013

- 23

^{x})

so I used the quotient rule to get:

(B+Ce

^{x})(1) - A(B+Ce

^{x})/(B+Ce

^{x})

^{2}

is this right so far? and if it is, how do I simplify it more?

- Thread starter coolbeans33
- Start date

- Thread starter
- #1

- Sep 19, 2013

- 23

so I used the quotient rule to get:

(B+Ce

is this right so far? and if it is, how do I simplify it more?

- Feb 5, 2012

- 1,621

Hi coolbeans,^{x})

so I used the quotient rule to get:

(B+Ce^{x})(1) - A(B+Ce^{x})/(B+Ce^{x})^{2}

is this right so far? and if it is, how do I simplify it more?

First we shall clarify a few doubts about your question. Are \(A,\, B\mbox{ and } C\) constants? Is your function the following?

\[f(x)=\frac{A}{B}+Ce^x\]

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- #3

- Sep 19, 2013

- 23

A, B, and C are all constants, and the function is A/(B + C*eHi coolbeans,

First we shall clarify a few doubts about your question. Are \(A,\, B\mbox{ and } C\) constants? Is your function the following?

\[f(x)=\frac{A}{B}+Ce^x\]

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- #5

- Sep 19, 2013

- 23

do I use the power rule or the chain rule first?I would write the function as:

\(\displaystyle f(x)=A\left(B+Ce^x \right)^{-1}\)

Now apply the power and chain rules.

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- #6

\(\displaystyle \frac{d}{dx}\left(\left(u(x) \right)^n \right)=n\cdot\left(u(x) \right)^{n-1}\cdot\frac{du}{dx}\)do I use the power rule or the chain rule first?

Which is just the Chain Rule...\(\displaystyle \frac{d}{dx}\left(\left(u(x) \right)^n \right)=n\cdot\left(u(x) \right)^{n-1}\cdot\frac{du}{dx}\)

- Mar 1, 2012

- 249

Your choice, it doesn't matter as long as you do bothdo I use the power rule or the chain rule first?

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- #9

- Sep 19, 2013

- 23

-1A(B+Ce

the part I don't get is how I'm supposed to take the derivative of the variables. do I just treat them like x and say they're equal to one?

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- #10

- Mar 5, 2012

- 9,774

First off, you're applying the chain rule, which has multiplication instead of addition.

-1A(B+Ce^{x})^{-2}+ d/dx (AB+ACe^{x})

the part I don't get is how I'm supposed to take the derivative of the variables. do I just treat them like x and say they're equal to one?

So you should have:

-1A(B+Ce

Then you have these A, B, and C, which are constants, not variables.

You should treat them the same as if they would read for instance 2, 3, respectively 4.

What would be d/dx (2*3+2*4*e

When you have that, you should replace any occurrences of 2, 3, and 4 again by A, B, and C respectively.

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- #11

- Sep 19, 2013

- 23

First off, you're applying the chain rule, which has multiplication instead of addition.

So you should have:

-1A(B+Ce^{x})^{-2}* d/dx (AB+ACe^{x})

Then you have these A, B, and C, which are constants, not variables.

You should treat them the same as if they would read for instance 2, 3, respectively 4.

What would be d/dx (2*3+2*4*e^{x})?

When you have that, you should replace any occurrences of 2, 3, and 4 again by A, B, and C respectively.

ok I just figured it out. just to make sure, if you multiply a constant (c) by x or e

Last edited:

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- #12

- Mar 5, 2012

- 9,774

If your constant is named C, then yes, you get Cx or Ceok I just figured it out. just to make sure, if you multiply a constant (c) by x or e^{x}, you're left with Cx or Ce^{x}right?

But if your constant is named c, then you would get cx or ce