What is (\partial)/(\partial x)((59)/((1+x^2+y^2))) ?

The answer to (\partial)/(\partial x)((59)/((1+x^2+y^2))) is -(118x)/((1+x^2+y^2)^2)

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(\partial)/(\partial x)((59)/((1+x^2+y^2)))

\frac{\partial}{\partial x} (\frac{59}{( 1 + x ^{2} + y ^{2} )})

- \frac{118 x}{( 1 + x ^{2} + y ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{59}{( 1 + x ^{2} + y ^{2} )})

Treat

y

as a constant

Take the constant out:

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

= 59 \frac{\partial}{\partial x} (\frac{1}{1 + x ^{2} + y ^{2}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 59 \frac{\partial}{\partial x} ((1 + x^{2} + y^{2})^{- 1})

Apply the chain rule:

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

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

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

= 59 (- \frac{1}{( 1 + x ^{2} + y ^{2} ) ^{2}} \cdot 2 x)

Simplify

59 (- \frac{1}{( 1 + x ^{2} + y ^{2} ) ^{2}} \cdot 2 x) : - \frac{118 x}{( 1 + x ^{2} + y ^{2} ) ^{2}}

= - \frac{118 x}{( 1 + x ^{2} + y ^{2} ) ^{2}}

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

(1 + x^{2}) y^{^{'}} + 4 x y = x, y (0) = 1

\int \frac{2}{( 1 - 2 x ) ^{2}} d x

\int_{- 1}^{0} e^{- x} d x

\frac{d}{d x} (5 tan (3 x))

What is (\partial)/(\partial x)((59)/((1+x^2+y^2))) ?
The answer to (\partial)/(\partial x)((59)/((1+x^2+y^2))) is -(118x)/((1+x^2+y^2)^2)