What is the derivative of f(w)=((2w^4-5w))/(9w) ?

The derivative of f(w)=((2w^4-5w))/(9w) is (2w^2)/3

What is the first derivative of f(w)=((2w^4-5w))/(9w) ?

The first derivative of f(w)=((2w^4-5w))/(9w) is (2w^2)/3

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derivative of

f (w) = \frac{( 2 w ^{4} - 5 w )}{9 w}

\frac{2 w ^{2}}{3}

Solution steps

\frac{d}{d w} (\frac{2 w ^{4} - 5 w}{9 w})

Take the constant out:

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

= \frac{1}{9} \frac{d}{d w} (\frac{2 w ^{4} - 5 w}{w})

Apply the Quotient Rule:

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

= \frac{1}{9} \cdot \frac{\frac{d}{d w} ( 2 w ^{4} - 5 w ) w - \frac{d w}{d w} ( 2 w ^{4} - 5 w )}{w ^{2}}

\frac{d}{d w} (2 w^{4} - 5 w) = 8 w^{3} - 5

\frac{d w}{d w} = 1

= \frac{1}{9} \cdot \frac{( 8 w ^{3} - 5 ) w - 1 \cdot ( 2 w ^{4} - 5 w )}{w ^{2}}

Simplify

\frac{1}{9} \cdot \frac{( 8 w ^{3} - 5 ) w - 1 \cdot ( 2 w ^{4} - 5 w )}{w ^{2}} : \frac{2 w ^{2}}{3}

= \frac{2 w ^{2}}{3}

\frac{d}{d x} (\frac{x ^{2} - 1}{x ^{2} + 1})

\int 2 t - 4 d t

f (t) = 2 e^{- t} cos (4 t) 1 (t)

\frac{d y}{d x} = (2 y + 1 cos (x))

\frac{d ^{2}}{d x ^{2}} (x^{3})

What is the derivative of f(w)=((2w^4-5w))/(9w) ?
The derivative of f(w)=((2w^4-5w))/(9w) is (2w^2)/3
What is the first derivative of f(w)=((2w^4-5w))/(9w) ?
The first derivative of f(w)=((2w^4-5w))/(9w) is (2w^2)/3