What is the derivative of ((3t^2+4))/(4t+4) ?

The derivative of ((3t^2+4))/(4t+4) is (3t^2+6t-4)/(4(t+1)^2)

What is the first derivative of ((3t^2+4))/(4t+4) ?

The first derivative of ((3t^2+4))/(4t+4) is (3t^2+6t-4)/(4(t+1)^2)

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derivative of

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

\frac{3 t ^{2} + 6 t - 4}{4 ( t + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d t} (3 t^{2} + 4) = 6 t

\frac{d}{d t} (4 t + 4) = 4

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

Simplify

\frac{6 t ( 4 t + 4 ) - 4 ( 3 t ^{2} + 4 )}{( 4 t + 4 ) ^{2}} : \frac{3 t ^{2} + 6 t - 4}{4 ( t + 1 ) ^{2}}

= \frac{3 t ^{2} + 6 t - 4}{4 ( t + 1 ) ^{2}}

What is the derivative of ((3t^2+4))/(4t+4) ?
The derivative of ((3t^2+4))/(4t+4) is (3t^2+6t-4)/(4(t+1)^2)
What is the first derivative of ((3t^2+4))/(4t+4) ?
The first derivative of ((3t^2+4))/(4t+4) is (3t^2+6t-4)/(4(t+1)^2)