basic calculus

can anyone answer these for me please

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

thanks

Hello buddy

Quote:

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Originally Posted by turbod15b

can anyone answer these for me please

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

thanks

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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

thanks man

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

cheers

Quote:

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Originally Posted by turbod15b

thanks man

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

cheers

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

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

Use the quotient rule to solve this one