What is the d/(dt)((t^3)/(4-3t+5)) ?

The d/(dt)((t^3)/(4-3t+5)) is (t^2(-2t+9))/(3(t-3)^2)

What is the first d/(dt)((t^3)/(4-3t+5)) ?

The first d/(dt)((t^3)/(4-3t+5)) is (t^2(-2t+9))/(3(t-3)^2)

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\frac{d}{d t} (\frac{t ^{3}}{4 - 3 t + 5})

\frac{t ^{2} ( - 2 t + 9 )}{3 ( t - 3 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{t ^{3}}{4 - 3 t + 5})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( t ^{3} ) ( 4 - 3 t + 5 ) - \frac{d}{d t} ( 4 - 3 t + 5 ) t ^{3}}{( 4 - 3 t + 5 ) ^{2}}

\frac{d}{d t} (t^{3}) = 3 t^{2}

\frac{d}{d t} (4 - 3 t + 5) = - 3

= \frac{3 t ^{2} ( 4 - 3 t + 5 ) - ( - 3 ) t ^{3}}{( 4 - 3 t + 5 ) ^{2}}

Simplify

\frac{3 t ^{2} ( 4 - 3 t + 5 ) - ( - 3 ) t ^{3}}{( 4 - 3 t + 5 ) ^{2}} : \frac{t ^{2} ( - 2 t + 9 )}{3 ( t - 3 ) ^{2}}

= \frac{t ^{2} ( - 2 t + 9 )}{3 ( t - 3 ) ^{2}}

x^{2} (1 - 2 x)

\frac{d}{d x} (\frac{1}{csc ( x )})

\int_{0}^{2} (2 x - x^{3}) d x

(7 P + 5)^{6}

What is the d/(dt)((t^3)/(4-3t+5)) ?
The d/(dt)((t^3)/(4-3t+5)) is (t^2(-2t+9))/(3(t-3)^2)
What is the first d/(dt)((t^3)/(4-3t+5)) ?
The first d/(dt)((t^3)/(4-3t+5)) is (t^2(-2t+9))/(3(t-3)^2)