What is the d/(db)(T/(-ln(b/a))) ?

The d/(db)(T/(-ln(b/a))) is T/(bln^2(b/a))

What is the first d/(db)(T/(-ln(b/a))) ?

The first d/(db)(T/(-ln(b/a))) is T/(bln^2(b/a))

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\frac{d}{d b} (\frac{T}{- ln ( \frac{b}{a} )})

\frac{T}{b ln ^{2} ( \frac{b}{a} )}

Solution steps

\frac{d}{d b} (\frac{T}{- ln ( \frac{b}{a} )})

Simplify

\frac{T}{- ln ( \frac{b}{a} )} : - \frac{T}{ln ( \frac{b}{a} )}

= \frac{d}{d b} (- \frac{T}{ln ( \frac{b}{a} )})

Treat

a, T

as a constant

Take the constant out:

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

= - T \frac{d}{d b} (\frac{1}{ln ( \frac{b}{a} )})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= - T \frac{d}{d b} (ln^{- 1} (\frac{b}{a}))

Apply the chain rule:

- \frac{1}{( ln ( \frac{b}{a} ) ) ^{2}} \frac{d}{d b} (ln (\frac{b}{a}))

= - \frac{1}{( ln ( \frac{b}{a} ) ) ^{2}} \frac{d}{d b} (ln (\frac{b}{a}))

\frac{d}{d b} (ln (\frac{b}{a})) = \frac{1}{b}

= - T (- \frac{1}{( ln ( \frac{b}{a} ) ) ^{2}} \cdot \frac{1}{b})

Simplify

- T (- \frac{1}{ln ^{2} ( \frac{b}{a} )} \cdot \frac{1}{b}) : \frac{T}{b ln ^{2} ( \frac{b}{a} )}

= \frac{T}{b ln ^{2} ( \frac{b}{a} )}

\frac{d}{d x} (x^{3} - 4 x^{2} + 1)

\frac{\partial}{\partial y} (2 y^{2} e^{x y^{2}})

\frac{d}{d x} (ln ((2 x + 1)^{5}))

7 cos (x)

What is the d/(db)(T/(-ln(b/a))) ?
The d/(db)(T/(-ln(b/a))) is T/(bln^2(b/a))
What is the first d/(db)(T/(-ln(b/a))) ?
The first d/(db)(T/(-ln(b/a))) is T/(bln^2(b/a))