What is the d/(dt)((e^{tb}-e^{ta})/(t(b-a))) ?

The d/(dt)((e^{tb}-e^{ta})/(t(b-a))) is (be^{bt}t-ae^{at}t-e^{bt}+e^{at})/(t^2(b-a))

What is the first d/(dt)((e^{tb}-e^{ta})/(t(b-a))) ?

The first d/(dt)((e^{tb}-e^{ta})/(t(b-a))) is (be^{bt}t-ae^{at}t-e^{bt}+e^{at})/(t^2(b-a))

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\frac{d}{d t} (\frac{e ^{t b} - e ^{t a}}{t ( b - a )})

\frac{b e ^{b t} t - a e ^{a t} t - e ^{b t} + e ^{a t}}{t ^{2} ( b - a )}

Solution steps

\frac{d}{d t} (\frac{e ^{t b} - e ^{t a}}{t ( b - a )})

Take the constant out:

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

= \frac{1}{b - a} \frac{d}{d t} (\frac{e ^{t b} - e ^{t a}}{t})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{b - a} \cdot \frac{\frac{d}{d t} ( e ^{t b} - e ^{t a} ) t - \frac{d t}{d t} ( e ^{t b} - e ^{t a} )}{t ^{2}}

\frac{d}{d t} (e^{t b} - e^{t a}) = e^{t b} b - e^{t a} a

\frac{d t}{d t} = 1

= \frac{1}{b - a} \cdot \frac{( e ^{t b} b - e ^{t a} a ) t - 1 \cdot ( e ^{t b} - e ^{t a} )}{t ^{2}}

Simplify

\frac{1}{b - a} \cdot \frac{( e ^{t b} b - e ^{t a} a ) t - 1 \cdot ( e ^{t b} - e ^{t a} )}{t ^{2}} : \frac{b e ^{b t} t - a e ^{a t} t - e ^{b t} + e ^{a t}}{t ^{2} ( b - a )}

= \frac{b e ^{b t} t - a e ^{a t} t - e ^{b t} + e ^{a t}}{t ^{2} ( b - a )}

y^{^{''}} + 9 y = 12 sin (3 t)

x \to 2 lim (\frac{x ^{3}}{3} - 7 x^{2})

f (y) = y^{4}

\frac{d ^{2}}{d x ^{2}} (sec (\frac{π}{4}))

\int sec (x) tan (x) - 9 e^{x} d x

What is the d/(dt)((e^{tb}-e^{ta})/(t(b-a))) ?
The d/(dt)((e^{tb}-e^{ta})/(t(b-a))) is (be^{bt}t-ae^{at}t-e^{bt}+e^{at})/(t^2(b-a))
What is the first d/(dt)((e^{tb}-e^{ta})/(t(b-a))) ?
The first d/(dt)((e^{tb}-e^{ta})/(t(b-a))) is (be^{bt}t-ae^{at}t-e^{bt}+e^{at})/(t^2(b-a))