Chain Rule

  • Thread starter FritoTaco
  • Start date
  • #1
133
23

Homework Statement


Hello,

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

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

Homework Equations



Product Rule: [itex]f(x)g'(x)+f'(x)g(x)[/itex]

Quotient Rule: [itex]\dfrac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}[/itex]

Chain Rule:

if [itex]y=g(f(x))[/itex]

then [itex]\dfrac{dy}{dx}=g'(f(x))\cdot f'(x)[/itex]

or

if [itex]y=g(u)[/itex] and [itex]u=f(x)[/itex]

then [itex]\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}[/itex]

The Attempt at a Solution



Quotient

[itex]y=\dfrac{6}{1+e^{-x}}[/itex]

[itex]y'=\dfrac{(1+e^{-x})(6)'-(1+e^{-x})'(6)}{(1+e^{-x})^2}[/itex]

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

[itex]y'=\dfrac{-(1+e^{-x})-6(-e^{-x})}{(1+e^{-x})^2}[/itex]

[itex]y'=\dfrac{-(1+e^{0})-6(-e^{0})}{(1+e^{0})^2}[/itex]

[itex]y'=1[/itex]

Book Answer:

[itex]y=\dfrac{6}{(1+e^{-x})}[/itex]

[itex]y'=\dfrac{(1+e^{-x})(0)-6(-e^{-x})}{(1+e^{-x})^2}[/itex]

[itex]y'=\dfrac{6(e^{-x})}{(1+e^{-x})^2}[/itex]

[itex]y'=\dfrac{6(e^{0})}{(1+e^{-x})^2}[/itex]

[itex]y'=\dfrac{6(1)}{(1+1)^2}[/itex]

[itex]y'=\dfrac{6}{2^2}[/itex]

[itex]y'=\dfrac{3}{2}[/itex]

[itex]y=\dfrac{3}{2}x+3[/itex]

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

  • #2
BvU
Science Advisor
Homework Helper
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Used chain rule here for ##-(1+e^{-x})##
There is no chain rule here: just a sum rule if you want: $$(1+e^{-x})' = 1' + (e^{-x})' = 0 - e^{-x} = - e^{-x} $$
 
  • #3
133
23
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.
 
  • #4
34,983
6,736

The Attempt at a Solution



Quotient

[itex]y=\dfrac{6}{1+e^{-x}}[/itex]

[itex]y'=\dfrac{(1+e^{-x})(6)'-(1+e^{-x})'(6)}{(1+e^{-x})^2}[/itex]

[itex]y'=\dfrac{(1+e^{-x})(0)-(1+e^{-x})(-e^{-x})(6)}{(1+e^{-x})^2}[/itex] Used chain rule here for [itex]-(1+e^{-x})[/itex]
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:
[itex]y'=\dfrac{-(1+e^{-x})-6(-e^{-x})}{(1+e^{-x})^2}[/itex]

[itex]y'=\dfrac{-(1+e^{0})-6(-e^{0})}{(1+e^{0})^2}[/itex]

[itex]y'=1[/itex]

Book Answer:

[itex]y=\dfrac{6}{(1+e^{-x})}[/itex]

[itex]y'=\dfrac{(1+e^{-x})(0)-6(-e^{-x})}{(1+e^{-x})^2}[/itex]

[itex]y'=\dfrac{6(e^{-x})}{(1+e^{-x})^2}[/itex]

[itex]y'=\dfrac{6(e^{0})}{(1+e^{-x})^2}[/itex]

[itex]y'=\dfrac{6(1)}{(1+1)^2}[/itex]

[itex]y'=\dfrac{6}{2^2}[/itex]

[itex]y'=\dfrac{3}{2}[/itex]

[itex]y=\dfrac{3}{2}x+3[/itex]

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

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