What is the d/(dy)((x)(2y^3+x-pi^7-4x^3+cos(3+2y)-5)) ?

The d/(dy)((x)(2y^3+x-pi^7-4x^3+cos(3+2y)-5)) is x(6y^2-2sin(2y+3))

What is the first d/(dy)((x)(2y^3+x-pi^7-4x^3+cos(3+2y)-5)) ?

The first d/(dy)((x)(2y^3+x-pi^7-4x^3+cos(3+2y)-5)) is x(6y^2-2sin(2y+3))

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\frac{d}{d y} ((x) (2 y^{3} + x - π^{7} - 4 x^{3} + cos (3 + 2 y) - 5))

x (6 y^{2} - 2 sin (2 y + 3))

Solution steps

\frac{d}{d y} ((x) (2 y^{3} + x - π^{7} - 4 x^{3} + cos (3 + 2 y) - 5))

Treat

x

as a constant

Take the constant out:

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

= x \frac{d}{d y} (2 y^{3} + x - π^{7} - 4 x^{3} + cos (3 + 2 y) - 5)

Apply the Sum/Difference Rule:

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

= x (\frac{d}{d y} (2 y^{3}) + \frac{d x}{d y} - \frac{d}{d y} (π^{7}) - \frac{d}{d y} (4 x^{3}) + \frac{d}{d y} (cos (3 + 2 y)) - \frac{d}{d y} (5))

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

\frac{d x}{d y} = 0

\frac{d}{d y} (π^{7}) = 0

\frac{d}{d y} (4 x^{3}) = 0

\frac{d}{d y} (cos (3 + 2 y)) = - sin (3 + 2 y) \cdot 2

\frac{d}{d y} (5) = 0

= x (6 y^{2} + 0 - 0 - 0 - sin (3 + 2 y) \cdot 2 - 0)

Simplify

= x (6 y^{2} - 2 sin (2 y + 3))

e^{(5 t - 10)} h (t - 2)

s + 5 ((s + 6) (s + 3))

3 x^{4} (- 4 - \cdot 4) + x^{0} - (- 7)

(t - 4) H (t - 2)

\frac{17 ( x ^{2} + y ^{2} )}{6 y + 37 x}

What is the d/(dy)((x)(2y^3+x-pi^7-4x^3+cos(3+2y)-5)) ?
The d/(dy)((x)(2y^3+x-pi^7-4x^3+cos(3+2y)-5)) is x(6y^2-2sin(2y+3))
What is the first d/(dy)((x)(2y^3+x-pi^7-4x^3+cos(3+2y)-5)) ?
The first d/(dy)((x)(2y^3+x-pi^7-4x^3+cos(3+2y)-5)) is x(6y^2-2sin(2y+3))