What is (\partial)/(\partial y)(log_{3}(x-y)) ?

The answer to (\partial)/(\partial y)(log_{3}(x-y)) is -1/(ln(3)(x-y))

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(\partial)/(\partial y)(log_{3}(x-y))

\frac{\partial}{\partial y} (lo g_{3} (x - y))

- \frac{1}{ln ( 3 ) ( x - y )}

Solution steps

\frac{\partial}{\partial y} (lo g_{3} (x - y))

Treat

x

as a constant

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{\partial}{\partial y} (\frac{ln ( x - y )}{ln ( 3 )})

Take the constant out:

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

= \frac{1}{ln ( 3 )} \frac{\partial}{\partial y} (ln (x - y))

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 3 )} \cdot \frac{1}{x - y} (- 1)

Simplify

\frac{1}{ln ( 3 )} \cdot \frac{1}{x - y} (- 1) : - \frac{1}{ln ( 3 ) ( x - y )}

= - \frac{1}{ln ( 3 ) ( x - y )}

\int \frac{( x ^{2} )}{x ^{2} + 6} d x

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = - 2, y^{^{'}} (0) = 4

cos^{2} (8 t)

y = cos (9 x)

\frac{d y}{d x}, y = (sec (x) + tan (x)) (sec (x) - tan (x))

What is (\partial)/(\partial y)(log_{3}(x-y)) ?
The answer to (\partial)/(\partial y)(log_{3}(x-y)) is -1/(ln(3)(x-y))