What is the d/(dy)((2*3/y-3y)(6-y*3/y)) ?

The d/(dy)((2*3/y-3y)(6-y*3/y)) is 3(-6/(y^2)-3)

What is the first d/(dy)((2*3/y-3y)(6-y*3/y)) ?

The first d/(dy)((2*3/y-3y)(6-y*3/y)) is 3(-6/(y^2)-3)

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\frac{d}{d y} ((2 \cdot \frac{3}{y} - 3 y) (6 - y \cdot \frac{3}{y}))

3 (- \frac{6}{y ^{2}} - 3)

Solution steps

\frac{d}{d y} ((2 \cdot \frac{3}{y} - 3 y) (6 - y \frac{3}{y}))

Simplify

(2 \cdot \frac{3}{y} - 3 y) (6 - y \frac{3}{y}) : 3 (\frac{6}{y} - 3 y)

= \frac{d}{d y} (3 (\frac{6}{y} - 3 y))

Take the constant out:

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

= 3 \frac{d}{d y} (\frac{6}{y} - 3 y)

Apply the Sum/Difference Rule:

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

= 3 (\frac{d}{d y} (\frac{6}{y}) - \frac{d}{d y} (3 y))

\frac{d}{d y} (\frac{6}{y}) = - \frac{6}{y ^{2}}

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

= 3 (- \frac{6}{y ^{2}} - 3)

x \to 6 lim (\frac{∣ 6 - x ∣}{6 - x})

\int_{0}^{\frac{π}{8}} tan (4 x) d x

\int arccos (1) d x

f (x) = \frac{8 x}{x ^{2} + 1}

y = x^{2} ln (x^{2})

What is the d/(dy)((2*3/y-3y)(6-y*3/y)) ?
The d/(dy)((2*3/y-3y)(6-y*3/y)) is 3(-6/(y^2)-3)
What is the first d/(dy)((2*3/y-3y)(6-y*3/y)) ?
The first d/(dy)((2*3/y-3y)(6-y*3/y)) is 3(-6/(y^2)-3)