Matemática, perguntado por lucasOprofissional, 1 ano atrás

Derive a seguinte função g(x) = 20.Sen²(x/2)

OBS.: Explique detalhadanente cada passo.

Soluções para a tarefa

Respondido por DanJR
1
Olá Lucas!

\\ \displaystyle \mathsf{g(x) = 20 \cdot \sin^2 \left ( \frac{x}{2} \right )} \\\\\\ \mathsf{g(x) = 20 \cdot \left [ \sin \left ( \frac{x}{2} \right ) \right ]^2} \\\\\\ \mathsf{g'(x) = 20 \cdot 2 \cdot \left [ \sin \left ( \frac{x}{2} \right ) \right ]^{2 - 1} \cdot \left [ \sin \left ( \frac{x}{2} \right ) \right ]'} \\\\\\ \mathsf{g'(x) = 20 \cdot 2 \cdot \left [ \sin \left ( \frac{x}{2} \right ) \right ]^1 \cdot \left [ \cos \left ( \frac{x}{2} \right ) \cdot \left ( \frac{x}{2} \right )' \right ]} \\\\\\ \mathsf{g'(x) = 20 \cdot 2 \cdot \sin \left ( \frac{x}{2} \right ) \cdot \left [ \cos \left ( \frac{x}{2} \right ) \cdot \left ( \frac{1}{2} \right ) \right ]} \\\\\\ \mathsf{g'(x) = 20 \cdot 2 \cdot \sin \left ( \frac{x}{2} \right ) \cdot \cos \left ( \frac{x}{2} \right ) \cdot \frac{1}{2}} \\\\\\ \mathsf{g'(x) = \frac{20 \cdot 2}{2} \cdot \sin \left ( \frac{x}{2} \right ) \cdot \cos \left ( \frac{x}{2} \right )} \\\\\\ \boxed{\mathsf{g'(x) = 20 \cdot \sin \left ( \frac{x}{2} \right ) \cdot \cos \left ( \frac{x}{2} \right )}}

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