What is (\partial)/(\partial x)(6x(x^2+3y^2-1)^2) ?

The answer to (\partial)/(\partial x)(6x(x^2+3y^2-1)^2) is 6(5x^4+18x^2y^2-6x^2+9y^4-6y^2+1)

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(\partial)/(\partial x)(6x(x^2+3y^2-1)^2)

\frac{\partial}{\partial x} (6 x (x^{2} + 3 y^{2} - 1)^{2})

6 (5 x^{4} + 18 x^{2} y^{2} - 6 x^{2} + 9 y^{4} - 6 y^{2} + 1)

Solution steps

\frac{\partial}{\partial x} (6 x (x^{2} + 3 y^{2} - 1)^{2})

Treat

y

as a constant

Take the constant out:

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

= 6 \frac{\partial}{\partial x} (x (x^{2} + 3 y^{2} - 1)^{2})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = (x^{2} + 3 y^{2} - 1)^{2}

= 6 (\frac{\partial}{\partial x} (x) (x^{2} + 3 y^{2} - 1)^{2} + \frac{\partial}{\partial x} ((x^{2} + 3 y^{2} - 1)^{2}) x)

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

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

= 6 (1 \cdot (x^{2} + 3 y^{2} - 1)^{2} + 4 x (x^{2} + 3 y^{2} - 1) x)

Simplify

6 (1 \cdot (x^{2} + 3 y^{2} - 1)^{2} + 4 x (x^{2} + 3 y^{2} - 1) x) : 6 (5 x^{4} + 18 x^{2} y^{2} - 6 x^{2} + 9 y^{4} - 6 y^{2} + 1)

= 6 (5 x^{4} + 18 x^{2} y^{2} - 6 x^{2} + 9 y^{4} - 6 y^{2} + 1)

n = 1 \sum \infty cos (n)

x \to 0 lim (\frac{cos ( x )}{x})

\frac{\partial}{\partial x} (x^{2} - y^{2} + 5 x y)

f (x) = 2 - x

y^{^{''}} - 4 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 9

What is (\partial)/(\partial x)(6x(x^2+3y^2-1)^2) ?
The answer to (\partial)/(\partial x)(6x(x^2+3y^2-1)^2) is 6(5x^4+18x^2y^2-6x^2+9y^4-6y^2+1)