1. ## basic calculus

thanks

2. Hello buddy

Originally Posted by turbod15b

thanks
a) = 3*1 = 3

b) = $\displaystyle 5*e^{5t}+(-2)e^{-2t} = 5e^{5t}-2e^{-2t}$

c) = 2x-5

d) = $\displaystyle \frac{d}{dt} sin(t)/cos(t) = \frac{1}{cos^2(t)}$

This is because

$\displaystyle \frac{d}{dt} sin(t)/cos(t) = sin(t)*cos^{-1}(t)$

$\displaystyle = cos(t)*cos(t)^{-1} + (-1)(-1) sin(x)*cos(x)^{-2}*sin(x)$

$\displaystyle = 1 + sin^2(t)cos^{2}(t) = 1+tan^2(x) = \frac{1}{cos^2(t)}$

Im lazy, i wrote t instead of theta. Sorry for that.

not sure, is$\displaystyle \frac{1}{cos^2(\theta)} = sec^2(\theta)$ ?

Cheers,
Rapha

3. thanks man

can u show me step by step how to get to the end

cheers

4. Originally Posted by turbod15b
thanks man

can u show me step by step how to get to the end

cheers
$\displaystyle tan(x) = \frac{sin(x)}{cos(x)}$

$\displaystyle \frac{d}{dx}tan(x) = \frac{d}{dx}\frac{sin(x)}{cos(x)}$

Use the quotient rule to solve this one