What is (\partial)/(\partial x)(aln(x)+bln(y)) ?

The answer to (\partial)/(\partial x)(aln(x)+bln(y)) is a/x

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(\partial)/(\partial x)(aln(x)+bln(y))

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

\frac{a}{x}

Solution steps

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

Treat

a, b, y

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (b ln (y)) = 0

= \frac{a}{x} + 0

Simplify

= \frac{a}{x}

(t - 2)^{3}

- \frac{x}{2}

\int \frac{( 3 )}{( u ^{2} - 1 )} d u

y = (x - 2)^{2} - 4, y = x + 1

\int \frac{cos ( \frac{π}{x} )}{x ^{2}} d x

What is (\partial)/(\partial x)(aln(x)+bln(y)) ?
The answer to (\partial)/(\partial x)(aln(x)+bln(y)) is a/x