MHB Derivative of a function with only variables

[coolbeans33]

I need to find the f'(x) when f(x)= A/B+C (e^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?

 

[Sudharaka]

I need to find the f'(x) when f(x)= A/B+C (e^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?

Hi coolbeans, :)

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

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

 

[coolbeans33]

Hi coolbeans, :)

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

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

A, B, and C are all constants, and the function is A/(B + C*e^x)

 

[MarkFL]

I would write the function as:

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

Now apply the power and chain rules. :D

 

[coolbeans33]

I would write the function as:

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

Now apply the power and chain rules. :D

do I use the power rule or the chain rule first?

 

[MarkFL]

do I use the power rule or the chain rule first?

$$\frac{d}{dx}\left(\left(u(x) \right)^n \right)=n\cdot\left(u(x) \right)^{n-1}\cdot\frac{du}{dx}$$

 

[Prove It]

$$\frac{d}{dx}\left(\left(u(x) \right)^n \right)=n\cdot\left(u(x) \right)^{n-1}\cdot\frac{du}{dx}$$

Which is just the Chain Rule...

 

[SuperSonic4]

do I use the power rule or the chain rule first?

Your choice, it doesn't matter as long as you do both

 

[coolbeans33]

this still makes no sense! after I applied the chain rule I got

-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?

 

[I like Serena]

this still makes no sense! after I applied the chain rule I got

-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?

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.

 

[coolbeans33]

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^x, you're left with Cx or Ce^x right?

 

[I like Serena]

ok 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?

If your constant is named C, then yes, you get Cx or Ce^x.
But if your constant is named c, then you would get cx or ce^x.