- #1
FritoTaco
- 132
- 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?