Find the equation of the tangent to the curve defined by y=ln(1+2^(-x)) at the point where x=0.
Help will be appreciated
Take the derivative of y and evaluate it at x = 0. This will be your slope value. To find the equation to the tangent, you use the formula:
$\displaystyle y - y_{1} = m(x - x_{1})$
where $\displaystyle (x_{1}, y_{1})$ is the point you're concerned with.
As for taking the derivative, make sure you're careful about using your chain rule! Show us what you've gotten down and we'll help you from there
Note that $\displaystyle 2^{-x}=e^{-xln(2)}$ if we take the derivative we get.
$\displaystyle \frac{d}{dx}2^{-x}=-ln(2)e^{-xln(2)}=-ln(2)2^{-x}
$
So now back to your derivative
$\displaystyle \frac{dy}{dx}=\frac{-ln(2)2^{-x}}{1+2^{-x}}$
evaluating at zero we get...
$\displaystyle \frac{dy}{dx}|_{x=0}=\frac{-ln(2)2^{-0}}{1+2^{-0}}=\frac{-ln(2)}{2}$
So i got
y` = 1 / (1+e^-x) * e^-x * (-1)
= - e^(-x) / (1+ e^-x)
mtangent @ x=0 = -1 / (1+1)
= -1/2
y - ln2 = 1/2(x-0)
2y - 2ln2 = x
x - 2y + 2ln2 = 0
Im not sure if that's right but looks fair enough. I would like any advice though.
I also have another question where it says: Find the equation of the tangent to the curve defined by y = e^x that is perpendicular to the line 3x + y = 1.
Thanks a lot in advance!