What is (\partial)/(\partial y)(y^x-ln(x^3y+6x)) ?

The answer to (\partial)/(\partial y)(y^x-ln(x^3y+6x)) is xy^{x-1}-(x^2)/(x^2y+6)

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(\partial)/(\partial y)(y^x-ln(x^3y+6x))

\frac{\partial}{\partial y} (y^{x} - ln (x^{3} y + 6 x))

x y^{x - 1} - \frac{x ^{2}}{x ^{2} y + 6}

Solution steps

\frac{\partial}{\partial y} (y^{x} - ln (x^{3} y + 6 x))

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (y^{x}) - \frac{\partial}{\partial y} (ln (x^{3} y + 6 x))

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

\frac{\partial}{\partial y} (ln (x^{3} y + 6 x)) = \frac{x ^{2}}{x ^{2} y + 6}

= x y^{x - 1} - \frac{x ^{2}}{x ^{2} y + 6}

x \to - 4 + lim (\frac{- 6 x}{x + 4})

\frac{\partial}{\partial x} (arctan (\frac{1}{x}))

\int e^{- s t} t sin (t) d t

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

x \to 1 + lim (\frac{∣ 1 - x ∣}{1 - x})

What is (\partial)/(\partial y)(y^x-ln(x^3y+6x)) ?
The answer to (\partial)/(\partial y)(y^x-ln(x^3y+6x)) is xy^{x-1}-(x^2)/(x^2y+6)