(\partial)/(\partial x)(1.015sqrt(E(m,x)bh^3m^{-1)L^{-4}})

解

\frac{\partial}{\partial x} (1.015 E (m, x) b h^{3} m^{- 1} L^{- 4})

1.015 \cdot \frac{1}{2 E ( m , x ) b h ^{3} m ^{- 1} L ^{- 4}} b h^{3} m^{- 1} L^{- 4} \frac{\partial}{\partial x} (E (m, x))

解答ステップ

\frac{\partial}{\partial x} (1.015 E (m, x) b h^{3} m^{- 1} L^{- 4})

b, h, L, m

を定数として扱う

定数を除去:

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

= 1.015 \frac{\partial}{\partial x} (E (m, x) b h^{3} m^{- 1} L^{- 4})

連鎖法則を適用:

\frac{1}{2 E ( m , x ) b h ^{3} m ^{- 1} L ^{- 4}} \frac{\partial}{\partial x} (E (m, x) b h^{3} m^{- 1} L^{- 4})

= \frac{1}{2 E ( m , x ) b h ^{3} m ^{- 1} L ^{- 4}} \frac{\partial}{\partial x} (E (m, x) b h^{3} m^{- 1} L^{- 4})

\frac{\partial}{\partial x} (E (m, x) b h^{3} m^{- 1} L^{- 4}) = b h^{3} m^{- 1} L^{- 4} \frac{\partial}{\partial x} (E (m, x))

= 1.015 \cdot \frac{1}{2 E ( m , x ) b h ^{3} m ^{- 1} L ^{- 4}} b h^{3} m^{- 1} L^{- 4} \frac{\partial}{\partial x} (E (m, x))