finding derivative of function with variables

[coolbeans33]

A, B, and C are all constants, and the function is A/(B + Ce^x). I simplified it to A(B+Ce^x)^-1 so I could apply the chain rule.

I applied the chain rule: -1A(B + Ce^x)^-2) * d/dx A(B + Ce^x)

I'm not sure if I did this right so far, but I'm left with having to take the derivative of a bunch of variables, and I don't even know how to do that

 

[DrPhil]

A, B, and C are all constants, and the function is A/(B + Ce^x). I simplified it to A(B+Ce^x)^-1 so I could apply the chain rule.

I applied the chain rule: -1A(B + Ce^x)^-2) * d/dx A(B + Ce^x)

I'm not sure if I did this right so far, but I'm left with having to take the derivative of a bunch of variables, and I don't even know how to do that

The derivative of a constant is 0, since it never changes value
.........d/dx (B + Ce^x) = C e^x

Note: the A is a constant multiplying the entire expression, and it can be moved outside. It will only appear once.

 

[HallsofIvy]

A, B, and C are all constants, and the function is A/(B + Ce^x). I simplified it to A(B+Ce^x)^-1 so I could apply the chain rule.

I applied the chain rule: -1A(B + Ce^x)^-2) * d/dx A(B + Ce^x)

The second "A" should not be there. it was \(\displaystyle B+ Ce^x\) that was to the "-1" power and so you only need to multiply by the derivative of that.

I'm not sure if I did this right so far, but I'm left with having to take the derivative of a bunch of variables, and I don't even know how to do that

The "bunch of variables" is \(\displaystyle B+ Ce^x\) which really has only the single variable, x. "B", "C", and "e" are constants. You surely should know that the derivative of "B+ Cf(x)" where B and C are constants, is C times the derivative of f(x). That is true because the derivative of "u(x)+ v(x)" is the derivative of u plus the derivative of v: (u+ v)'= u'+ v'. You should also know the "product rule, (u(x)v(x))'= u'v+ uv'. Of course, if one of those is a constant, say, u= C, then u'= 0 so that reduces to (Cv(x))'= 0v(x)+ Cv'= Cv'(x).