Question

.Find the derivative of

cos x

with respect to “

x

” from first principles.​

Answer 1

$\frac{d\cos(x)}{dx} \\ y = \cos(x) \\ \frac{dy}{dx} = \frac{y2 - y1}{x2 - x1} \\ dy = y2 - y1 \: (limiting \: delta \: x \: tends \: to \: zero) \\ dx = x2 - x1 \\ delta \: x + x1 = x2 \\ same \: as \: y \: \\ delta \: y + y1 = y2 \\ \frac{dy}{dx} = \frac{d \cos(x + delta \: x) - \cos(x) }{delta \: x + x1 - x1} \\ \cos(a + b) = \cos(a) \cos(b) - \sin(a) \sin(b) \\ \frac{d \cos(x) \cos(delta \: x) - \sin(x) \sin(delta \: x )- \cos(x) }{delta \: x} \\ \\ \frac{ \cos(x) - \sin(x) (delta \: x) - \cos(x) }{delta \: x } \: \: \: \: \: \: \: because (\cos(delta \: x) = 1 \: \: (sin(delta \: x) = delta \: x\\ \frac{ \ - sin(x) (delta \: x)}{delta \: x} = - \sin(x) \\ \frac{d \cos(x) }{dx} = - \sin(x)$

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