What is the derivative of g(t)=(3t^2-1)/(2t+5) ?

The derivative of g(t)=(3t^2-1)/(2t+5) is (6t^2+30t+2)/((2t+5)^2)

What is the first derivative of g(t)=(3t^2-1)/(2t+5) ?

The first derivative of g(t)=(3t^2-1)/(2t+5) is (6t^2+30t+2)/((2t+5)^2)

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derivative of

g (t) = \frac{3 t ^{2} - 1}{2 t + 5}

\frac{6 t ^{2} + 30 t + 2}{( 2 t + 5 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{3 t ^{2} - 1}{2 t + 5})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( 3 t ^{2} - 1 ) ( 2 t + 5 ) - \frac{d}{d t} ( 2 t + 5 ) ( 3 t ^{2} - 1 )}{( 2 t + 5 ) ^{2}}

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

\frac{d}{d t} (2 t + 5) = 2

= \frac{6 t ( 2 t + 5 ) - 2 ( 3 t ^{2} - 1 )}{( 2 t + 5 ) ^{2}}

Expand

6 t (2 t + 5) - 2 (3 t^{2} - 1) : 6 t^{2} + 30 t + 2

= \frac{6 t ^{2} + 30 t + 2}{( 2 t + 5 ) ^{2}}

What is the derivative of g(t)=(3t^2-1)/(2t+5) ?
The derivative of g(t)=(3t^2-1)/(2t+5) is (6t^2+30t+2)/((2t+5)^2)
What is the first derivative of g(t)=(3t^2-1)/(2t+5) ?
The first derivative of g(t)=(3t^2-1)/(2t+5) is (6t^2+30t+2)/((2t+5)^2)