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Derivative of a function with only variables

coolbeans33

New member
Sep 19, 2013
23
I need to find the f'(x) when f(x)= A/B+C (ex)

so I used the quotient rule to get:

(B+Cex)(1) - A(B+Cex)/(B+Cex)2

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

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I need to find the f'(x) when f(x)= A/B+C (ex)

so I used the quotient rule to get:

(B+Cex)(1) - A(B+Cex)/(B+Cex)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

New member
Sep 19, 2013
23
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*ex)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would write the function as:

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

Now apply the power and chain rules. :D
 

coolbeans33

New member
Sep 19, 2013
23
I would write the function as:

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

Now apply the power and chain rules. :D
do I use the power rule or the chain rule first?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
do I use the power rule or the chain rule first?
\(\displaystyle \frac{d}{dx}\left(\left(u(x) \right)^n \right)=n\cdot\left(u(x) \right)^{n-1}\cdot\frac{du}{dx}\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
\(\displaystyle \frac{d}{dx}\left(\left(u(x) \right)^n \right)=n\cdot\left(u(x) \right)^{n-1}\cdot\frac{du}{dx}\)
Which is just the Chain Rule...
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249

coolbeans33

New member
Sep 19, 2013
23
this still makes no sense! after I applied the chain rule I got

-1A(B+Cex)-2 + d/dx (AB+ACex)

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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
this still makes no sense! after I applied the chain rule I got

-1A(B+Cex)-2 + d/dx (AB+ACex)

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+Cex)-2 * d/dx (AB+ACex)

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

coolbeans33

New member
Sep 19, 2013
23
First off, you're applying the chain rule, which has multiplication instead of addition.
So you should have:

-1A(B+Cex)-2 * d/dx (AB+ACex)

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*ex)?
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 ex, you're left with Cx or Cex right?
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
ok I just figured it out. just to make sure, if you multiply a constant (c) by x or ex, you're left with Cx or Cex right?
If your constant is named C, then yes, you get Cx or Cex.
But if your constant is named c, then you would get cx or cex.