Does anyone know how I get from
y=2 tan-1 (et)
to
dy/dx = sech t ?
(the minus one after tan and the t following the e are powers!!)
dy/dx = 0 since there's no x in the expression.
To find dy/dt, use the chain rule:
$\displaystyle \frac{dy}{dt} = \frac{2 e^t}{1 + (e^t)^2} = \frac{2 e^t}{1 + e^{2t}}$.
Now divide numertaor and denominator by $\displaystyle e^t$:
$\displaystyle \frac{dy}{dt} = \frac{2}{e^{-t} + e^{t}} = \text{sech}\, t$ by definition.
$\displaystyle y=2arctan(e^t)$
solve for e^t (not really necessary, but I figure you'll see it better this way)
$\displaystyle tan(y/2)=e^t$
take the derivative
$\displaystyle \frac 12 sec^2(y/2) ~y\prime=e^t$
solve for y'
$\displaystyle y\prime=2~cos^2(y/2) ~e^t$
substitute for y from the first equation
$\displaystyle y\prime=2~cos^2(arctan(e^t)) ~e^t$
simplify (draw a triangle if necessary... or just memorize the derivative of arctangent and save yourself this extra work
$\displaystyle y\prime=\frac{2e^t}{e^{2t}+1}$
divide numerator and denominator by e^t
$\displaystyle y\prime=\frac{2}{e^t+\frac{1}{e^t}}$
$\displaystyle y\prime=\frac{2}{e^t+e^{-t}}$
an equation is equal to 1 over it's reciprocal
$\displaystyle y\prime=\frac {~~~~~1~~~~~}{\frac{e^t+e^{-t}}{2}}$
the denominator is the equation for cosh(t)
$\displaystyle y\prime=\frac {1}{cosh(t)}$
and simplify
$\displaystyle y\prime=sech(t)$
sech(x) = 1 / cosh(x)
We know that the derivative of cosh(x) is sinh(x).
The derivative of something like 1/f(x) is -f'(x)/f²(x)
---> the derivative of sech(x) is $\displaystyle \frac{-\sinh(x)}{\cosh^2(x)}=\frac{-1}{\cosh(x)} \cdot \tanh(x)$
You can take a look at these explanations, here : Hyperbolic Secant -- from Wolfram MathWorld
Using the same process I used before
$\displaystyle y=arctan(x)$
$\displaystyle tan(y) = x$
differentiate
$\displaystyle sec^2(y) y\prime = 1$
$\displaystyle y\prime = cos^2(y)$
substitute from the first equation
$\displaystyle y\prime = cos^2(arctan(x))$
$\displaystyle y\prime = \left(\frac 1{\sqrt{x^2+1}}\right)^2$
$\displaystyle y\prime = \frac 1{x^2+1}$
and because y = arctan(x)
$\displaystyle \frac d{dx} ~arctan(x) = \frac 1{x^2+1}$