What is (\partial)/(\partial y)(-1/(ln(10)(x-y))) ?

The answer to (\partial)/(\partial y)(-1/(ln(10)(x-y))) is -1/(ln(10)(x-y)^2)

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(\partial)/(\partial y)(-1/(ln(10)(x-y)))

\frac{\partial}{\partial y} (- \frac{1}{ln ( 10 ) ( x - y )})

- \frac{1}{ln ( 10 ) ( x - y ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (- \frac{1}{ln ( 10 ) ( x - y )})

Treat

x

as a constant

Take the constant out:

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

= - \frac{1}{ln ( 10 )} \frac{\partial}{\partial y} (\frac{1}{x - y})

Apply exponent rule:

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

= - \frac{1}{ln ( 10 )} \frac{\partial}{\partial y} ((x - y)^{- 1})

Apply the chain rule:

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

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

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

= - \frac{1}{ln ( 10 )} (- \frac{1}{( x - y ) ^{2}} (- 1))

Simplify

- \frac{1}{ln ( 10 )} (- \frac{1}{( x - y ) ^{2}} (- 1)) : - \frac{1}{ln ( 10 ) ( x - y ) ^{2}}

= - \frac{1}{ln ( 10 ) ( x - y ) ^{2}}

f (x) = \frac{8}{sin ( x ) + cos ( x )}

f (x) = x^{2} - 12, g (x) = x - 6, [23, 6]

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

f (x) = 1 + cos (x), at x = \frac{3 π}{2}

\int x^{4} e^{- 2 x} d x

What is (\partial)/(\partial y)(-1/(ln(10)(x-y))) ?
The answer to (\partial)/(\partial y)(-1/(ln(10)(x-y))) is -1/(ln(10)(x-y)^2)