What is (\partial)/(\partial x)((10x)/(1+x^2-y)) ?

The answer to (\partial)/(\partial x)((10x)/(1+x^2-y)) is (10(-x^2-y+1))/((1+x^2-y)^2)

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(\partial)/(\partial x)((10x)/(1+x^2-y))

\frac{\partial}{\partial x} (\frac{10 x}{1 + x ^{2} - y})

\frac{10 ( - x ^{2} - y + 1 )}{( 1 + x ^{2} - y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{10 x}{1 + x ^{2} - y})

Treat

y

as a constant

Take the constant out:

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

= 10 \frac{\partial}{\partial x} (\frac{x}{1 + x ^{2} - y})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 10 \cdot \frac{\frac{\partial}{\partial x} ( x ) ( 1 + x ^{2} - y ) - \frac{\partial}{\partial x} ( 1 + x ^{2} - y ) x}{( 1 + x ^{2} - y ) ^{2}}

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

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

= 10 \cdot \frac{1 \cdot ( 1 + x ^{2} - y ) - 2 xx}{( 1 + x ^{2} - y ) ^{2}}

Simplify

10 \cdot \frac{1 \cdot ( 1 + x ^{2} - y ) - 2 xx}{( 1 + x ^{2} - y ) ^{2}} : \frac{10 ( - x ^{2} - y + 1 )}{( 1 + x ^{2} - y ) ^{2}}

= \frac{10 ( - x ^{2} - y + 1 )}{( 1 + x ^{2} - y ) ^{2}}

\frac{\partial}{\partial y} (\frac{y}{2})

f (x) = 3 x^{2} - 1

\frac{d y}{d x} = x + 6 y, y (0) = 2

f (x) = \frac{4 x + 5}{x - 1}, (2, 13)

2 x, 2 x

What is (\partial)/(\partial x)((10x)/(1+x^2-y)) ?
The answer to (\partial)/(\partial x)((10x)/(1+x^2-y)) is (10(-x^2-y+1))/((1+x^2-y)^2)