Question

JAAAAAAAAAAAAAAAAAAGTE RHO .... SO GIVE ME ALL THE DERIVATIVES OF ALLL THE TRIGONOMETRIC VALUES ..PZZZZ

Answer 1

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Answer 2

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$\frac{d}{dx}[sin\: x] = cos\: x$
$\frac{d}{dx}[sin\: u] = cos\: u × \frac{du}{dx}$

$\frac{d}{dx}[cos\: x] = -sin\: x$
$\frac{d}{dx}[sin\: u] = -sin\: u × \frac{du}{dx}$

$\frac{d}{dx}[tan\: x] = sec^{2}\: x$
$\frac{d}{dx}[tan\: u] = sec^{2}\: u × \frac{du}{dx}$

$\frac{d}{dx}[cot\: x] = -cosec^{2}\: x$
$\frac{d}{dx}[tan\: u] = -cosec^{2}\: u × \frac{du}{dx}$

$\frac{d}{dx}[sec\: x] = sec\: x\: tan\:x$
$\frac{d}{dx}[sec\: u] = sec\: u\: tan\:u × \frac{du}{dx}$

$\frac{d}{dx}[cosec\: x] = -cosec\: x\: cot\:x$
$\frac{d}{dx}[cosec\: u] = -cosec\: u\: cot\:u × \frac{du}{dx}$