What is (\partial)/(\partial x)(3sqrt(xy^2-2xy+z)) ?

The answer to (\partial)/(\partial x)(3sqrt(xy^2-2xy+z)) is (3(y^2-2y))/(2sqrt(xy^2-2xy+z))

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(\partial)/(\partial x)(3sqrt(xy^2-2xy+z))

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

\frac{3 ( y ^{2} - 2 y )}{2 x y ^{2} - 2 x y + z}

Solution steps

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

Treat

y, z

as constants

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= 3 \cdot \frac{1}{2 x y ^{2} - 2 x y + z} (y^{2} - 2 y)

Simplify

3 \cdot \frac{1}{2 x y ^{2} - 2 x y + z} (y^{2} - 2 y) : \frac{3 ( y ^{2} - 2 y )}{2 x y ^{2} - 2 x y + z}

= \frac{3 ( y ^{2} - 2 y )}{2 x y ^{2} - 2 x y + z}

y^{^{''}} - 4 y^{^{'}} + 7 y = 0

\int_{- 2}^{3} \frac{28}{x ^{4}} d x

x^{2} = y^{2}, (1, - 1)

f (x) = x^{2} - 7 x + 7, at x = 0

\frac{\partial}{\partial x} (- y \cdot cos (x))

What is (\partial)/(\partial x)(3sqrt(xy^2-2xy+z)) ?
The answer to (\partial)/(\partial x)(3sqrt(xy^2-2xy+z)) is (3(y^2-2y))/(2sqrt(xy^2-2xy+z))