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

Lösung

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

\frac{d}{d T} (E (r, T)) (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 (r, T)

Schritte zur Lösung

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

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

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

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

= \frac{d}{d T} (E (r, T)) (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 (r, T)

e x p an d \frac{x ^{3} - 3 x ^{2}}{( x - 1 ) ^{3}}

e x p an d (lo g_{10} (x)) (x + 1) (x + 2) (x + 3)

e x p an d \frac{x ( x - y ln ( x ) )}{y ( y - x ln ( y ) )}

e x p an d 7 (10 + 3)

e x p an d \frac{1}{s ( 6 s + 1 )}