How do you find the derivative of $s=cost/(t-1)$ ?

Question

2032 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-30T13:51:40

$(ds)/dt=-(t+cost-sint)/(t-1)^2$

$s=cost/(t-1)$

Use the quotient rule to differentiate $s$ :

$d/dx((p(x))/(q(x)))=(q(x)p'(x)-p(x)q'(x))/([q(x)]^2)$

Let $p(x)=cost$ and $q(x)=t-1$

Then $p'(x)=-sint$ and $q'(x)=1$ and $[q(x)]^2=(t-1)^2$

$(ds)/dt=(-(t-1)sint-cost)/(t-1)^2=-(tsint+cost-sint)/(t-1)^2$

How do you find the derivative of s=cost/(t-1)s=cost/(t-1)?