Question

Calcular a derivada f ( θ ) = 2 cos (2θ 2 − 3θ + 1 )

Answer 1

Imagino que a função seja essa:

$f(\theta)=2cos(2 \theta^2-3 \theta+1)\\\\g(\theta) =2 \theta^2-3 \theta+1 \\\\f(g) = 2cos(g)$

Com a regra da cadeia, observando que, pela mudanca acima:

$f \rightarrow g \rightarrow \theta$

Entao:

$\dfrac{df}{d \theta} = \dfrac{df}{dg} \dfrac{dg}{d \theta}\\\\\dfrac{df}{dg} = -2sin(g)\\\dfrac{dg}{d \theta} = 4 \theta - 3$

$\dfrac{df}{d \theta} = 2 \dfrac{df}{dg} \dfrac{dg}{d \theta}\\\\\dfrac{df}{dg} = -sin(g)\\\dfrac{dg}{d \theta} = 4 \theta - 3\\\\\dfrac{df}{d \theta} = -2sin(2 \theta^2-3 \theta+1)(4 \theta - 3)$