d/(dr(T))(({E}(T,b))(10-10T× b^{-0.14r(T)}))

Lösung

\frac{d ^{2}}{d r ( T ) d T} ((E (T, b)) (10 - 10 T \cdot b^{- 0.14 r (T)}))

\frac{d}{d T} (E (T, b)) (0 - (- 1.4 T b^{- 0.14 r (T)} ln (b))) + 1.4 ln (b) (- 0.14 b^{- 0.14 r (T)} T ln (b) \frac{d}{d T} (r (T)) + b^{- 0.14 r (T)}) E (T, b)

Schritte zur Lösung

\frac{d ^{2}}{d r ( T ) d T} ((E (T, b)) (10 - 10 T b^{- 0.14 r (T)}))

\frac{d}{d r ( T )} (E (T, b) (10 - 10 T b^{- 0.14 r (T)})) = E (T, b) (0 - (- 1.4 T b^{- 0.14 r (T)} ln (b)))

= \frac{d}{d T} (E (T, b) (0 - (- 1.4 T b^{- 0.14 r (T)} ln (b))))

\frac{d}{d T} (E (T, b) (0 - (- 1.4 T b^{- 0.14 r (T)} ln (b)))) = \frac{d}{d T} (E (T, b)) (0 - (- 1.4 T b^{- 0.14 r (T)} ln (b))) + 1.4 ln (b) (- 0.14 b^{- 0.14 r (T)} T ln (b) \frac{d}{d T} (r (T)) + b^{- 0.14 r (T)}) E (T, b)

= \frac{d}{d T} (E (T, b)) (0 - (- 1.4 T b^{- 0.14 r (T)} ln (b))) + 1.4 ln (b) (- 0.14 b^{- 0.14 r (T)} T ln (b) \frac{d}{d T} (r (T)) + b^{- 0.14 r (T)}) E (T, b)

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