1. ## Differential of tan

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!!)

2. Originally Posted by bbbabybel
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:

$\frac{dy}{dt} = \frac{2 e^t}{1 + (e^t)^2} = \frac{2 e^t}{1 + e^{2t}}$.

Now divide numertaor and denominator by $e^t$:

$\frac{dy}{dt} = \frac{2}{e^{-t} + e^{t}} = \text{sech}\, t$ by definition.

3. Originally Posted by bbbabybel
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!!)
$y=2arctan(e^t)$

solve for e^t (not really necessary, but I figure you'll see it better this way)

$tan(y/2)=e^t$

take the derivative

$\frac 12 sec^2(y/2) ~y\prime=e^t$

solve for y'

$y\prime=2~cos^2(y/2) ~e^t$

substitute for y from the first equation

$y\prime=2~cos^2(arctan(e^t)) ~e^t$

simplify (draw a triangle if necessary... or just memorize the derivative of arctangent and save yourself this extra work

$y\prime=\frac{2e^t}{e^{2t}+1}$

divide numerator and denominator by e^t

$y\prime=\frac{2}{e^t+\frac{1}{e^t}}$

$y\prime=\frac{2}{e^t+e^{-t}}$

an equation is equal to 1 over it's reciprocal

$y\prime=\frac {~~~~~1~~~~~}{\frac{e^t+e^{-t}}{2}}$

the denominator is the equation for cosh(t)

$y\prime=\frac {1}{cosh(t)}$

and simplify

$y\prime=sech(t)$

4. Right...thanks to both. The product rule cleared it up and I did mean dy/dt haha. Can someone just clarify what the differential of tan-1 x is? Because I think that's where I got lost!

5. Or the chain rule rather....oh dear,I'm trying!

6. Originally Posted by bbbabybel
Right...thanks to both. The product rule cleared it up and I did mean dy/dt haha. Can someone just clarify what the differential of tan-1 x is? Because I think that's where I got lost!
Hello,

$(\tan^{-1} t)'=\frac{1}{1+t^2}$

Thus, using the chain rule :

$(\tan^{-1} f(x))'=\frac{f'(x)}{1+f^2(x)}$

7. While we're on the lovely subject, what's the differential of sech x?

8. Originally Posted by bbbabybel
While we're on the lovely subject, what's the differential of sech x?

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 $\frac{-\sinh(x)}{\cosh^2(x)}=\frac{-1}{\cosh(x)} \cdot \tanh(x)$

You can take a look at these explanations, here : Hyperbolic Secant -- from Wolfram MathWorld

9. Originally Posted by bbbabybel
Right...thanks to both. The product rule cleared it up and I did mean dy/dt haha. Can someone just clarify what the differential of tan-1 x is? Because I think that's where I got lost!
Using the same process I used before

$y=arctan(x)$

$tan(y) = x$

differentiate

$sec^2(y) y\prime = 1$

$y\prime = cos^2(y)$

substitute from the first equation

$y\prime = cos^2(arctan(x))$

$y\prime = \left(\frac 1{\sqrt{x^2+1}}\right)^2$

$y\prime = \frac 1{x^2+1}$

and because y = arctan(x)

$\frac d{dx} ~arctan(x) = \frac 1{x^2+1}$