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

The derivative of ((t+5)2)/(3(2t^2-3)3) is (2(-2t^2-20t-3))/(9(2t^2-3)^2)

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

The first derivative of ((t+5)2)/(3(2t^2-3)3) is (2(-2t^2-20t-3))/(9(2t^2-3)^2)

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derivative of

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

\frac{2 ( - 2 t ^{2} - 20 t - 3 )}{9 ( 2 t ^{2} - 3 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d}{d t} (t + 5) = 1

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

= \frac{2}{3 \cdot 3} \cdot \frac{1 \cdot ( 2 t ^{2} - 3 ) - 4 t ( t + 5 )}{( 2 t ^{2} - 3 ) ^{2}}

Simplify

\frac{2}{3 \cdot 3} \cdot \frac{1 \cdot ( 2 t ^{2} - 3 ) - 4 t ( t + 5 )}{( 2 t ^{2} - 3 ) ^{2}} : \frac{2 ( - 2 t ^{2} - 20 t - 3 )}{9 ( 2 t ^{2} - 3 ) ^{2}}

= \frac{2 ( - 2 t ^{2} - 20 t - 3 )}{9 ( 2 t ^{2} - 3 ) ^{2}}

\int 2 x - 9 d x

\frac{d y}{d r} = 2 + y - 2 r + 3

\frac{d}{d x} (sin^{3} (x) + cos^{3} (x))

\frac{d}{d x} (x^{3} + y^{3} - 9 x y)

What is the derivative of ((t+5)2)/(3(2t^2-3)3) ?
The derivative of ((t+5)2)/(3(2t^2-3)3) is (2(-2t^2-20t-3))/(9(2t^2-3)^2)
What is the first derivative of ((t+5)2)/(3(2t^2-3)3) ?
The first derivative of ((t+5)2)/(3(2t^2-3)3) is (2(-2t^2-20t-3))/(9(2t^2-3)^2)