What is (\partial)/(\partial x)(ln(axy)+b^{xy}) ?

The answer to (\partial)/(\partial x)(ln(axy)+b^{xy}) is 1/x+yb^{yx}ln(b)

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(\partial)/(\partial x)(ln(axy)+b^{xy})

\frac{\partial}{\partial x} (ln (a x y) + b^{x y})

\frac{1}{x} + y b^{y x} ln (b)

Solution steps

\frac{\partial}{\partial x} (ln (a x y) + b^{x y})

Treat

a, y, b

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= \frac{1}{x} + y b^{y x} ln (b)

220 + 11 x - 11 x^{2}, 11 x^{2} - 55 x

x^{5} e^{x}

\int 2 \cdot (x - 6)^{3} d x

\frac{d y}{d x} = 7 x^{6} e^{- y}

f (x) = 7 + 4 x^{2} - 2 x^{3}, at x = a

What is (\partial)/(\partial x)(ln(axy)+b^{xy}) ?
The answer to (\partial)/(\partial x)(ln(axy)+b^{xy}) is 1/x+yb^{yx}ln(b)