What is (\partial)/(\partial x)(-ylog_{10}(x)-(1-y)log_{10}(1-x)) ?

The answer to (\partial)/(\partial x)(-ylog_{10}(x)-(1-y)log_{10}(1-x)) is (y-x)/(ln(10)x(x-1))

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(\partial)/(\partial x)(-ylog_{10}(x)-(1-y)log_{10}(1-x))

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial x} (y lo g_{10} (x)) = \frac{y}{ln ( 10 ) x}

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

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

Simplify

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

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

\int cos^{5} (x) d x

x \to 9 lim (cos (\frac{π x}{4}))

y \to \infty lim (\frac{2 y ^{2} - 3 y + 5}{y ^{2} - 5 y + 2})

x \to 2 lim (x^{3} + 3 x^{2})

f (x) = 8 x^{2} - x^{3}

What is (\partial)/(\partial x)(-ylog_{10}(x)-(1-y)log_{10}(1-x)) ?
The answer to (\partial)/(\partial x)(-ylog_{10}(x)-(1-y)log_{10}(1-x)) is (y-x)/(ln(10)x(x-1))