How do you differentiate $s = \frac{1 + sin t}{1 + cos t}$ $s=(1+sint)/(1+cost)$ ?

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How do you differentiate $s = \frac{1 + sin t}{1 + cos t}$ $s=(1+sint)/(1+cost)$ ?

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Answer 1 · 2017-02-27T19:01:06

$\frac{d s}{d t} = \frac{1 + cos t + sin t}{{(1 + cos t)}^{2}}$ $(ds)/(dt) = ( 1+ cost + sint)/(1+cost)^2$

Explanation:

Using the quotient rule:

$\frac{d s}{d t} = \frac{(1 + cos t) \frac{d}{d t} (1 + sin t) - (1 + sin t) \frac{d}{d t} (1 + cos t)}{{(1 + cos t)}^{2}}$ $(ds)/(dt) = ( (1+cost) d/(dt) (1+sint) - (1+sint) d/(dt) (1+cost))/(1+cost)^2$

$\frac{d s}{d t} = \frac{(1 + cos t) cos t + (1 + sin t) sin t}{{(1 + cos t)}^{2}}$ $(ds)/(dt) = ( (1+cost) cost + (1+sint) sint)/(1+cost)^2$

$\frac{d s}{d t} = \frac{cos t + {cos}^{2} t + sin t + {sin}^{2} t}{{(1 + cos t)}^{2}}$ $(ds)/(dt) = ( cost+cos^2t + sint+sin^2t)/(1+cost)^2$

$\frac{d s}{d t} = \frac{1 + cos t + sin t}{{(1 + cos t)}^{2}}$ $(ds)/(dt) = ( 1+ cost + sint)/(1+cost)^2$

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How do you differentiate $s = \frac{1 + sin t}{1 + cos t}$ $s=(1+sint)/(1+cost)$ ?

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