# Chain Rule

## Homework Statement

Hello,

I need help finding the derivative. The question wants me to find the equation of the tangent line to the curve

$y=\dfrac{6}{1+e^{-x}}$ at point $(0, 3)$. I'm unclear on when to use the chain rule at certain areas.

## Homework Equations

Product Rule: $f(x)g'(x)+f'(x)g(x)$

Quotient Rule: $\dfrac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}$

Chain Rule:

if $y=g(f(x))$

then $\dfrac{dy}{dx}=g'(f(x))\cdot f'(x)$

or

if $y=g(u)$ and $u=f(x)$

then $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$

## The Attempt at a Solution

Quotient

$y=\dfrac{6}{1+e^{-x}}$

$y'=\dfrac{(1+e^{-x})(6)'-(1+e^{-x})'(6)}{(1+e^{-x})^2}$

$y'=\dfrac{(1+e^{-x})(0)-(1+e^{-x})(-e^{-x})(6)}{(1+e^{-x})^2}$ Used chain rule here for $-(1+e^{-x})$

$y'=\dfrac{-(1+e^{-x})-6(-e^{-x})}{(1+e^{-x})^2}$

$y'=\dfrac{-(1+e^{0})-6(-e^{0})}{(1+e^{0})^2}$

$y'=1$

$y=\dfrac{6}{(1+e^{-x})}$

$y'=\dfrac{(1+e^{-x})(0)-6(-e^{-x})}{(1+e^{-x})^2}$

$y'=\dfrac{6(e^{-x})}{(1+e^{-x})^2}$

$y'=\dfrac{6(e^{0})}{(1+e^{-x})^2}$

$y'=\dfrac{6(1)}{(1+1)^2}$

$y'=\dfrac{6}{2^2}$

$y'=\dfrac{3}{2}$

$y=\dfrac{3}{2}x+3$

Everything goes terribly wrong for me when I use the chain rule. Also, at then end, why didn't they square the denominator?

## Answers and Replies

BvU
Homework Helper
Used chain rule here for ##-(1+e^{-x})##
There is no chain rule here: just a sum rule if you want: $$(1+e^{-x})' = 1' + (e^{-x})' = 0 - e^{-x} = - e^{-x}$$

FritoTaco
There is no chain rule here: just a sum rule if you want: $$(1+e^{-x})' = 1' + (e^{-x})' = 0 - e^{-x} = - e^{-x}$$

Thanks, I didn't see that little detail. That helped.

Mark44
Mentor

## The Attempt at a Solution

Quotient

$y=\dfrac{6}{1+e^{-x}}$

$y'=\dfrac{(1+e^{-x})(6)'-(1+e^{-x})'(6)}{(1+e^{-x})^2}$

$y'=\dfrac{(1+e^{-x})(0)-(1+e^{-x})(-e^{-x})(6)}{(1+e^{-x})^2}$ Used chain rule here for $-(1+e^{-x})$
I would never use the quotient rule for a function where the numerator is a constant. I would write ##\frac 6 {1 + e^{-x}}## as ##6(1 + e^{-x})^{-1}## and use the constant multiple rule followed by the chain rule.

Having said that, your work in the line above is incorrect. You should NOT use the chain rule on ##1 + e^{-x}##, at least not at first. The derivative of ##1 + e^{-x}## is simply ##-e^{-x}##.

##\frac d {dx} \left( 1 + e^{-x} \right) = \frac d {dx} 1 + \frac d {dx} e^{-x} = 0 - e^{-x}##
I used the sum rule followed by the chain rule.
FritoTaco said:
$y'=\dfrac{-(1+e^{-x})-6(-e^{-x})}{(1+e^{-x})^2}$

$y'=\dfrac{-(1+e^{0})-6(-e^{0})}{(1+e^{0})^2}$

$y'=1$

$y=\dfrac{6}{(1+e^{-x})}$

$y'=\dfrac{(1+e^{-x})(0)-6(-e^{-x})}{(1+e^{-x})^2}$

$y'=\dfrac{6(e^{-x})}{(1+e^{-x})^2}$

$y'=\dfrac{6(e^{0})}{(1+e^{-x})^2}$

$y'=\dfrac{6(1)}{(1+1)^2}$

$y'=\dfrac{6}{2^2}$

$y'=\dfrac{3}{2}$

$y=\dfrac{3}{2}x+3$

Everything goes terribly wrong for me when I use the chain rule. Also, at then end, why didn't they square the denominator?

FritoTaco