What is the derivative of v(t)=(4t^2+2)/(5t+5) ?

The derivative of v(t)=(4t^2+2)/(5t+5) is (2(2t^2+4t-1))/(5(t+1)^2)

What is the first derivative of v(t)=(4t^2+2)/(5t+5) ?

The first derivative of v(t)=(4t^2+2)/(5t+5) is (2(2t^2+4t-1))/(5(t+1)^2)

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d t} (4 t^{2} + 2) = 8 t

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

= \frac{8 t ( 5 t + 5 ) - 5 ( 4 t ^{2} + 2 )}{( 5 t + 5 ) ^{2}}

Simplify

\frac{8 t ( 5 t + 5 ) - 5 ( 4 t ^{2} + 2 )}{( 5 t + 5 ) ^{2}} : \frac{2 ( 2 t ^{2} + 4 t - 1 )}{5 ( t + 1 ) ^{2}}

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

\int (x + 2) 2 - x d x

y^{^{'}} = k y (1 - y)

\frac{d}{d x} (\frac{1}{x} + ln (x))

5 x^{2} - x^{3}

f (x) = \frac{1}{x ^{2} + 4}

What is the derivative of v(t)=(4t^2+2)/(5t+5) ?
The derivative of v(t)=(4t^2+2)/(5t+5) is (2(2t^2+4t-1))/(5(t+1)^2)
What is the first derivative of v(t)=(4t^2+2)/(5t+5) ?
The first derivative of v(t)=(4t^2+2)/(5t+5) is (2(2t^2+4t-1))/(5(t+1)^2)