What is the d/(dy)((y^6)/6+1/(16y^4)) ?

The d/(dy)((y^6)/6+1/(16y^4)) is y^5-1/(4y^5)

What is the first d/(dy)((y^6)/6+1/(16y^4)) ?

The first d/(dy)((y^6)/6+1/(16y^4)) is y^5-1/(4y^5)

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\frac{d}{d y} (\frac{y ^{6}}{6} + \frac{1}{16 y ^{4}})

y^{5} - \frac{1}{4 y ^{5}}

Solution steps

\frac{d}{d y} (\frac{y ^{6}}{6} + \frac{1}{16 y ^{4}})

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (\frac{y ^{6}}{6}) + \frac{d}{d y} (\frac{1}{16 y ^{4}})

\frac{d}{d y} (\frac{y ^{6}}{6}) = y^{5}

\frac{d}{d y} (\frac{1}{16 y ^{4}}) = - \frac{1}{4 y ^{5}}

= y^{5} - \frac{1}{4 y ^{5}}

3 + t^{5} + sin (π t)

\frac{d y}{d x} = \frac{y}{x} - cos (\frac{y}{x})

\int (x^{4} - 6 x) d x

\int \frac{sin ( 20 x )}{1 + cos ^{2} ( 20 x )} d x

What is the d/(dy)((y^6)/6+1/(16y^4)) ?
The d/(dy)((y^6)/6+1/(16y^4)) is y^5-1/(4y^5)
What is the first d/(dy)((y^6)/6+1/(16y^4)) ?
The first d/(dy)((y^6)/6+1/(16y^4)) is y^5-1/(4y^5)