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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d y}{d t} + a y = 0

2 y^{^{'}} + x y^{^{''}} = 0

\frac{\partial}{\partial z} (2 x)

\int 4 x sec^{2} (2 x) d x

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