# basic calculus

• May 18th 2009, 09:28 AM
turbod15b
basic calculus

http://img527.imageshack.us/img527/360/dsc01414dqo.jpg

thanks
• May 18th 2009, 09:35 AM
Rapha
Hello buddy

Quote:

Originally Posted by turbod15b

http://img527.imageshack.us/img527/360/dsc01414dqo.jpg

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
• May 18th 2009, 09:38 AM
turbod15b
thanks man

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

cheers
• May 18th 2009, 09:40 AM
e^(i*pi)
Quote:

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