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}}

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)