What is (\partial)/(\partial x)((3x^2+5x+1)*(y+4)) ?

The answer to (\partial)/(\partial x)((3x^2+5x+1)*(y+4)) is (y+4)(6x+5)

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(\partial)/(\partial x)((3x^2+5x+1)*(y+4))

\frac{\partial}{\partial x} ((3 x^{2} + 5 x + 1) \cdot (y + 4))

(y + 4) (6 x + 5)

Solution steps

\frac{\partial}{\partial x} ((3 x^{2} + 5 x + 1) (y + 4))

Treat

y

as a constant

Take the constant out:

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

= (y + 4) \frac{\partial}{\partial x} (3 x^{2} + 5 x + 1)

Apply the Sum/Difference Rule:

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

= (y + 4) (\frac{\partial}{\partial x} (3 x^{2}) + \frac{\partial}{\partial x} (5 x) + \frac{\partial}{\partial x} (1))

\frac{\partial}{\partial x} (3 x^{2}) = 6 x

\frac{\partial}{\partial x} (5 x) = 5

\frac{\partial}{\partial x} (1) = 0

= (y + 4) (6 x + 5 + 0)

Simplify

= (y + 4) (6 x + 5)

\frac{d y}{d x} = y (x y^{3} - 1)

\frac{d}{d x} (arccsc (\frac{x + 1}{1 - x}))

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e - 4 x, y (0) = - 5, y^{^{'}} (0) = 1

\int 11 - 10 x - x^{2} d x

\int \frac{2}{( 1 + x ^{2} )} - \frac{4}{x} d x

What is (\partial)/(\partial x)((3x^2+5x+1)*(y+4)) ?
The answer to (\partial)/(\partial x)((3x^2+5x+1)*(y+4)) is (y+4)(6x+5)