What is (\partial)/(\partial x)(t/(x-1)) ?

The answer to (\partial)/(\partial x)(t/(x-1)) is -t/((x-1)^2)

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(\partial)/(\partial x)(t/(x-1))

\frac{\partial}{\partial x} (\frac{t}{x - 1})

- \frac{t}{( x - 1 ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{t}{x - 1})

Treat

t

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

= t (- \frac{1}{( x - 1 ) ^{2}} \cdot 1)

Simplify

t (- \frac{1}{( x - 1 ) ^{2}} \cdot 1) : - \frac{t}{( x - 1 ) ^{2}}

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

\int 2 x - 2 y d x

\frac{1}{1 - z ^{2}}, 0

\frac{\partial}{\partial x} (\frac{1}{2} ln (x))

(x^{2} + 1) y^{^{'}} + 8 x (y - 1) = 0, y (0) = 3

\frac{\partial}{\partial x} (4 x^{2} - 4 x y + x yz)

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