What is the d/(dt)(((t^3-6t^2+7))/((2t^2+6t))) ?

The d/(dt)(((t^3-6t^2+7))/((2t^2+6t))) is (t^4+6t^3-18t^2-14t-21)/(2t^2(t+3)^2)

What is the first d/(dt)(((t^3-6t^2+7))/((2t^2+6t))) ?

The first d/(dt)(((t^3-6t^2+7))/((2t^2+6t))) is (t^4+6t^3-18t^2-14t-21)/(2t^2(t+3)^2)

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\frac{d}{d t} (\frac{( t ^{3} - 6 t ^{2} + 7 )}{( 2 t ^{2} + 6 t )})

\frac{t ^{4} + 6 t ^{3} - 18 t ^{2} - 14 t - 21}{2 t ^{2} ( t + 3 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{( t ^{3} - 6 t ^{2} + 7 )}{( 2 t ^{2} + 6 t )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( t ^{3} - 6 t ^{2} + 7 ) ( 2 t ^{2} + 6 t ) - \frac{d}{d t} ( 2 t ^{2} + 6 t ) ( t ^{3} - 6 t ^{2} + 7 )}{( 2 t ^{2} + 6 t ) ^{2}}

\frac{d}{d t} (t^{3} - 6 t^{2} + 7) = 3 t^{2} - 12 t

\frac{d}{d t} (2 t^{2} + 6 t) = 4 t + 6

= \frac{( 3 t ^{2} - 12 t ) ( 2 t ^{2} + 6 t ) - ( 4 t + 6 ) ( t ^{3} - 6 t ^{2} + 7 )}{( 2 t ^{2} + 6 t ) ^{2}}

Simplify

\frac{( 3 t ^{2} - 12 t ) ( 2 t ^{2} + 6 t ) - ( 4 t + 6 ) ( t ^{3} - 6 t ^{2} + 7 )}{( 2 t ^{2} + 6 t ) ^{2}} : \frac{t ^{4} + 6 t ^{3} - 18 t ^{2} - 14 t - 21}{2 t ^{2} ( t + 3 ) ^{2}}

= \frac{t ^{4} + 6 t ^{3} - 18 t ^{2} - 14 t - 21}{2 t ^{2} ( t + 3 ) ^{2}}

e^{3 x + 2} + 5

(((e^{e})^{e})^{x})^{2}

5 x^{3} + 16 b^{2}

y = \frac{( x + 2 ) ^{2}}{x}

f (\frac{1}{x ^{4}})

What is the d/(dt)(((t^3-6t^2+7))/((2t^2+6t))) ?
The d/(dt)(((t^3-6t^2+7))/((2t^2+6t))) is (t^4+6t^3-18t^2-14t-21)/(2t^2(t+3)^2)
What is the first d/(dt)(((t^3-6t^2+7))/((2t^2+6t))) ?
The first d/(dt)(((t^3-6t^2+7))/((2t^2+6t))) is (t^4+6t^3-18t^2-14t-21)/(2t^2(t+3)^2)