# Derivative of a function with only variables

#### coolbeans33

##### New member
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
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
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

Staff member
I would write the function as:

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

Now apply the power and chain rules. #### coolbeans33

##### New member
I would write the function as:

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

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

#### MarkFL

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

#### SuperSonic4

##### Well-known member
MHB Math Helper
do I use the power rule or the chain rule first?
Your choice, it doesn't matter as long as you do both

#### coolbeans33

##### New member
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
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
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?

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.