What is (\partial)/(\partial x)(3(x-1)2+2(y+3)2+7) ?

The answer to (\partial)/(\partial x)(3(x-1)2+2(y+3)2+7) is 6

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(\partial)/(\partial x)(3(x-1)2+2(y+3)2+7)

\frac{\partial}{\partial x} (3 (x - 1) 2 + 2 (y + 3) 2 + 7)

6

Solution steps

\frac{\partial}{\partial x} (3 (x - 1) \cdot 2 + 2 (y + 3) \cdot 2 + 7)

Simplify

3 (x - 1) \cdot 2 + 2 (y + 3) \cdot 2 + 7 : 6 x + 4 y + 13

= \frac{\partial}{\partial x} (6 x + 4 y + 13)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (6 x) + \frac{\partial}{\partial x} (4 y) + \frac{\partial}{\partial x} (13)

\frac{\partial}{\partial x} (6 x) = 6

\frac{\partial}{\partial x} (4 y) = 0

\frac{\partial}{\partial x} (13) = 0

= 6 + 0 + 0

Simplify

= 6

\int (e^{- 5 x}) d x

x y^{2} \frac{d y}{d x} = x^{3} + y^{3}

\int_{\frac{π}{2}}^{π} 5 cot (\frac{x}{2}) d x

\int - \frac{2 x ^{3} + 24 x}{x ^{4} - 16} d x

(1 + y^{2}) d x = x d y

What is (\partial)/(\partial x)(3(x-1)2+2(y+3)2+7) ?
The answer to (\partial)/(\partial x)(3(x-1)2+2(y+3)2+7) is 6