What is the derivative of y=(w^4+5w^2+w)/(w^2) ?

The derivative of y=(w^4+5w^2+w)/(w^2) is (\frac{dw)/(dy)(2w^3-1)}{w^2}

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

The first derivative of y=(w^4+5w^2+w)/(w^2) is (\frac{dw)/(dy)(2w^3-1)}{w^2}

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derivative of

y = \frac{w ^{4} + 5 w ^{2} + w}{w ^{2}}

\frac{\frac{d w}{d y} ( 2 w ^{3} - 1 )}{w ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d y} (w^{4} + 5 w^{2} + w) = 4 w^{3} \frac{d w}{d y} + 10 w \frac{d w}{d y} + \frac{d w}{d y}

\frac{d}{d y} (w^{2}) = 2 w \frac{d w}{d y}

= \frac{( 4 w ^{3} \frac{d w}{d y} + 10 w \frac{d w}{d y} + \frac{d w}{d y} ) w ^{2} - 2 w \frac{d w}{d y} ( w ^{4} + 5 w ^{2} + w )}{( w ^{2} ) ^{2}}

Simplify

\frac{( 4 w ^{3} \frac{d w}{d y} + 10 w \frac{d w}{d y} + \frac{d w}{d y} ) w ^{2} - 2 w \frac{d w}{d y} ( w ^{4} + 5 w ^{2} + w )}{( w ^{2} ) ^{2}} : \frac{\frac{d w}{d y} ( 2 w ^{3} - 1 )}{w ^{2}}

= \frac{\frac{d w}{d y} ( 2 w ^{3} - 1 )}{w ^{2}}

y = \frac{2}{x}, y = 8 x, y = \frac{1}{2} x

\int (\frac{1}{x} - \frac{5}{4 x ^{2} + 1}) d x

\int \frac{( 17 )}{( x - 1 ) ( x ^{2} + 16 )} d x

\int x (3 x^{2} - 2) d x

\frac{d}{d x} (\frac{1}{5} x^{5} - 3 x^{4} + 9 x^{2} + 1)

What is the derivative of y=(w^4+5w^2+w)/(w^2) ?
The derivative of y=(w^4+5w^2+w)/(w^2) is (\frac{dw)/(dy)(2w^3-1)}{w^2}
What is the first derivative of y=(w^4+5w^2+w)/(w^2) ?
The first derivative of y=(w^4+5w^2+w)/(w^2) is (\frac{dw)/(dy)(2w^3-1)}{w^2}