d/(dθ)((h)(y(h,θ)-tan(θ)x(h,θ)+1))

解

\frac{d}{d θ} ((h) (y (h, θ) - tan (θ) x (h, θ) + 1))

h (\frac{d}{d θ} (y (h, θ)) - (sec^{2} (θ) x (h, θ) + \frac{d}{d θ} (x (h, θ)) tan (θ)) + 0)

解答ステップ

\frac{d}{d θ} ((h) (y (h, θ) - tan (θ) x (h, θ) + 1))

h

を定数として扱う

定数を除去:

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

= h \frac{d}{d θ} (y (h, θ) - tan (θ) x (h, θ) + 1)

和/差の法則を適用:

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

= h (\frac{d}{d θ} (y (h, θ)) - \frac{d}{d θ} (tan (θ) x (h, θ)) + \frac{d}{d θ} (1))

\frac{d}{d θ} (tan (θ) x (h, θ)) = sec^{2} (θ) x (h, θ) + \frac{d}{d θ} (x (h, θ)) tan (θ)

\frac{d}{d θ} (1) = 0

= h (\frac{d}{d θ} (y (h, θ)) - (sec^{2} (θ) x (h, θ) + \frac{d}{d θ} (x (h, θ)) tan (θ)) + 0)