d/(dθ)((A)(5pir(θ)^2-4r(θ)^2θ-4r(θ)^2sin(θ)))

解

\frac{d}{d θ} ((A) (5 π r (θ) (θ)^{2} - 4 r (θ) (θ)^{2} θ - 4 r (θ) (θ)^{2} sin (θ)))

A (5 π (2 θ r (θ) + \frac{d}{d θ} (r (θ)) θ^{2}) - 4 (3 θ^{2} r (θ) + \frac{d}{d θ} (r (θ)) θ^{3}) - 4 (2 θ sin (θ) r (θ) + (cos (θ) r (θ) + \frac{d}{d θ} (r (θ)) sin (θ)) θ^{2}))

解答ステップ

\frac{d}{d θ} ((A) (5 π r (θ) (θ)^{2} - 4 r (θ) (θ)^{2} θ - 4 r (θ) (θ)^{2} sin (θ)))

簡素化

(A) (5 π r (θ) (θ)^{2} - 4 r (θ) (θ)^{2} θ - 4 r (θ) (θ)^{2} sin (θ)) : A (5 π θ^{2} r (θ) - 4 θ^{3} r (θ) - 4 θ^{2} sin (θ) r (θ))

= \frac{d}{d θ} (A (5 π θ^{2} r (θ) - 4 θ^{3} r (θ) - 4 θ^{2} sin (θ) r (θ)))

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= A \frac{d}{d θ} (5 π θ^{2} r (θ) - 4 θ^{3} r (θ) - 4 θ^{2} sin (θ) r (θ))

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= A (\frac{d}{d θ} (5 π θ^{2} r (θ)) - \frac{d}{d θ} (4 θ^{3} r (θ)) - \frac{d}{d θ} (4 θ^{2} sin (θ) r (θ)))

\frac{d}{d θ} (5 π θ^{2} r (θ)) = 5 π (2 θ r (θ) + \frac{d}{d θ} (r (θ)) θ^{2})

\frac{d}{d θ} (4 θ^{3} r (θ)) = 4 (3 θ^{2} r (θ) + \frac{d}{d θ} (r (θ)) θ^{3})

\frac{d}{d θ} (4 θ^{2} sin (θ) r (θ)) = 4 (2 θ sin (θ) r (θ) + (cos (θ) r (θ) + \frac{d}{d θ} (r (θ)) sin (θ)) θ^{2})

= A (5 π (2 θ r (θ) + \frac{d}{d θ} (r (θ)) θ^{2}) - 4 (3 θ^{2} r (θ) + \frac{d}{d θ} (r (θ)) θ^{3}) - 4 (2 θ sin (θ) r (θ) + (cos (θ) r (θ) + \frac{d}{d θ} (r (θ)) sin (θ)) θ^{2}))