What is (\partial)/(\partial x)(y^x-ln(x^4y+6x)) ?

The answer to (\partial)/(\partial x)(y^x-ln(x^4y+6x)) is y^xln(y)-(4yx^3+6)/(yx^4+6x)

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(\partial)/(\partial x)(y^x-ln(x^4y+6x))

\frac{\partial}{\partial x} (y^{x} - ln (x^{4} y + 6 x))

y^{x} ln (y) - \frac{4 y x ^{3} + 6}{y x ^{4} + 6 x}

Solution steps

\frac{\partial}{\partial x} (y^{x} - ln (x^{4} y + 6 x))

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (y^{x}) - \frac{\partial}{\partial x} (ln (x^{4} y + 6 x))

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

\frac{\partial}{\partial x} (ln (x^{4} y + 6 x)) = \frac{4 y x ^{3} + 6}{y x ^{4} + 6 x}

= y^{x} ln (y) - \frac{4 y x ^{3} + 6}{y x ^{4} + 6 x}

What is (\partial)/(\partial x)(y^x-ln(x^4y+6x)) ?
The answer to (\partial)/(\partial x)(y^x-ln(x^4y+6x)) is y^xln(y)-(4yx^3+6)/(yx^4+6x)