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

The answer to (\partial)/(\partial y)(x^2+(y-1)^2+z^2) is 2(y-1)

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

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

2 (y - 1)

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (x^{2}) = 0

\frac{\partial}{\partial y} ((y - 1)^{2}) = 2 (y - 1)

\frac{\partial}{\partial y} (z^{2}) = 0

= 0 + 2 (y - 1) + 0

Simplify

= 2 (y - 1)

x \to 4 lim (\frac{∣ x + 4 ∣}{x + 4})

\int \frac{5 x ^{2} + 8 x + 5}{( x - 5 ) ( x ^{2} + 9 )} d x

f (x) = 8 x sin (x)

\int_{1}^{9} \frac{x ^{2} - 1}{x} d x

f (x) = x, g (x) = - x, x = 1, x = 4

What is (\partial)/(\partial y)(x^2+(y-1)^2+z^2) ?
The answer to (\partial)/(\partial y)(x^2+(y-1)^2+z^2) is 2(y-1)