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)

(\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)

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

\int \frac{5 x ^{2} - 10 x}{4 3 x ^{3}} d x

d er i v a t i v e 1 - 49 x^{2} arccos (7 x)

x \to 2 lim (7 (x) + 10)

f (x) = x^{\frac{4}{5}}

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)