What is the d/(dM)((N)(10^A10^{-bM})) ?

The d/(dM)((N)(10^A10^{-bM})) is -ln(10)*10^{A-bM}Nb

What is the first d/(dM)((N)(10^A10^{-bM})) ?

The first d/(dM)((N)(10^A10^{-bM})) is -ln(10)*10^{A-bM}Nb

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\frac{d}{d M} ((N) (1 0^{A} 1 0^{- b M}))

- ln (10) \cdot 1 0^{A - b M} N b

Solution steps

\frac{d}{d M} ((N) (1 0^{A} \cdot 1 0^{- b M}))

Simplify

(N) (1 0^{A} \cdot 1 0^{- b M}) : 1 0^{A - b M} N

= \frac{d}{d M} (1 0^{A - b M} N)

Take the constant out:

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

= N \frac{d}{d M} (1 0^{A - b M})

Apply exponent rule:

a^{b} = e^{b l n (a)}

1 0^{A - b M} = e^{(A - b M) l n (10)}

= N \frac{d}{d M} (e^{(A - b M) l n (10)})

Apply the chain rule:

e^{(A - b M) l n (10)} \frac{d}{d M} ((A - b M) ln (10))

= e^{(A - b M) l n (10)} \frac{d}{d M} ((A - b M) ln (10))

\frac{d}{d M} ((A - b M) ln (10)) = - ln (10) b

= N e^{(A - b M) l n (10)} (- ln (10) b)

Simplify

N e^{(A - b M) l n (10)} (- ln (10) b) : - ln (10) \cdot 1 0^{A - b M} N b

= - ln (10) \cdot 1 0^{A - b M} N b

(\frac{( 5 ^{2} ) ^{- 2}}{15})^{- 2}

(- 0.5)^{3} \cdot (- 2)^{3}

- 2 e^{(1)^{2} - (- 1)^{2}} 2^{2} (- 1)

e^{3} \cdot e^{6}

e^{- \frac{1}{2} l n (t^{2} - 1)}

What is the d/(dM)((N)(10^A10^{-bM})) ?
The d/(dM)((N)(10^A10^{-bM})) is -ln(10)*10^{A-bM}Nb
What is the first d/(dM)((N)(10^A10^{-bM})) ?
The first d/(dM)((N)(10^A10^{-bM})) is -ln(10)*10^{A-bM}Nb