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

The derivative of (t^5-3t^2)/(2t) is 1/2 (4t^3-3)

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

The first derivative of (t^5-3t^2)/(2t) is 1/2 (4t^3-3)

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derivative of

\frac{t ^{5} - 3 t ^{2}}{2 t}

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d}{d t} (t^{5} - 3 t^{2}) = 5 t^{4} - 6 t

\frac{d t}{d t} = 1

= \frac{1}{2} \cdot \frac{( 5 t ^{4} - 6 t ) t - 1 \cdot ( t ^{5} - 3 t ^{2} )}{t ^{2}}

\frac{( 5 t ^{4} - 6 t ) t - 1 \cdot ( t ^{5} - 3 t ^{2} )}{t ^{2}} = 4 t^{3} - 3

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

\int \frac{x}{x + 5} d x

\int (x - 2) (3 x - 4) d x

\int 3 y sin (x y) d x

\frac{d y}{d x} = 7 x^{5} + 6

f (x) = 7 e^{- x} - 4 e^{- 7 x}

What is the derivative of (t^5-3t^2)/(2t) ?
The derivative of (t^5-3t^2)/(2t) is 1/2 (4t^3-3)
What is the first derivative of (t^5-3t^2)/(2t) ?
The first derivative of (t^5-3t^2)/(2t) is 1/2 (4t^3-3)