What is the derivative of (y^4)/4+1/(8y^2) ?

The derivative of (y^4)/4+1/(8y^2) is y^3-1/(4y^3)

What is the first derivative of (y^4)/4+1/(8y^2) ?

The first derivative of (y^4)/4+1/(8y^2) is y^3-1/(4y^3)

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derivative of

\frac{y ^{4}}{4} + \frac{1}{8 y ^{2}}

y^{3} - \frac{1}{4 y ^{3}}

Solution steps

\frac{d}{d y} (\frac{y ^{4}}{4} + \frac{1}{8 y ^{2}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d y} (\frac{y ^{4}}{4}) + \frac{d}{d y} (\frac{1}{8 y ^{2}})

\frac{d}{d y} (\frac{y ^{4}}{4}) = y^{3}

\frac{d}{d y} (\frac{1}{8 y ^{2}}) = - \frac{1}{4 y ^{3}}

= y^{3} - \frac{1}{4 y ^{3}}

\int_{1}^{4} - 9 e^{x} d x

(- 3, 7), (2, - 6)

\int 4 x e^{6 x} d x

- \frac{1}{2} x^{2} + \frac{5}{2} x - 2, [- 2, 7]

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

What is the derivative of (y^4)/4+1/(8y^2) ?
The derivative of (y^4)/4+1/(8y^2) is y^3-1/(4y^3)
What is the first derivative of (y^4)/4+1/(8y^2) ?
The first derivative of (y^4)/4+1/(8y^2) is y^3-1/(4y^3)