Question

A derivada de g(x)=4cos(√x) é:

Answer 1

$\sf \: g(x) = 4 \cos \Big(\sqrt{x}\Big) \\ \\ \sf \: g'(x) = \frac{d}{dx} \Bigg(4 \cos\Big( \sqrt{x} \Big) \Bigg) \\ \\ \sf \: g' (x) = 4 \times \frac{d}{dx} \Bigg( \cos\Big( \sqrt{x}\Big) \Bigg) \\ \\ \sf \: g'(x) = 4 \times \frac{d}{dg} \Big( \cos(g) \Big) \times \frac{d}{dx} \Big( \sqrt{x} \Big) \\ \\ \sf \: g'(x) = 4 \times \Bigg( \sin(g) \times \frac{d}{dx}\Big( \sqrt{x} \Big) \Bigg) \\ \\ \sf \: g'(x) = 4 \times \Bigg( - \sin\Big( \sqrt{x} \Big) \times \frac{1}{2 \sqrt{x} } \Bigg) \\ \\ \sf \: 4 \times \Bigg( - \frac{ \sin\Big( \sqrt{x}\Big) }{2 \sqrt{x} } \Bigg) \\ \\ \sf \: - 4 \times \frac{ \sin\Big( \sqrt{x} \Big) }{2 \sqrt{x} } \\ \\ \sf \: - 2 \times \frac{ \sin\Big( \sqrt{x} \Big) }{ \sqrt{x} } \\ \\ \boxed{\sf \: - \frac{2 \sin\Big( \sqrt{x} \Big) }{ \sqrt{x} }}$

Att: Nerd1990