What is (\partial)/(\partial x)(arctan(y/((x-1)))) ?

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

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(\partial)/(\partial x)(arctan(y/((x-1))))

\frac{\partial}{\partial x} (arctan (\frac{y}{( x - 1 )}))

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

Solution steps

\frac{\partial}{\partial x} (arctan (\frac{y}{( x - 1 )}))

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (62 - 2 x^{2} - 2 y^{2})

\frac{d}{d x} (ln (\frac{1}{1 + x}))

f (x) = arcsec (x)

y^{^{'}} + 7 x e^{y} = 0

\int_{- 1}^{0} 7 x d x

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