can anyone answer these for me please
thanks
Hello buddy
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