What is (\partial)/(\partial x)(aln(a(x^2+y^2))) ?

The answer to (\partial)/(\partial x)(aln(a(x^2+y^2))) is (2ax)/(x^2+y^2)

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(\partial)/(\partial x)(aln(a(x^2+y^2)))

\frac{\partial}{\partial x} (a ln (a (x^{2} + y^{2})))

\frac{2 a x}{x ^{2} + y ^{2}}

Solution steps

\frac{\partial}{\partial x} (a ln (a (x^{2} + y^{2})))

Treat

a, y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

\frac{1}{a ( x ^{2} + y ^{2} )} \frac{\partial}{\partial x} (a (x^{2} + y^{2}))

= \frac{1}{a ( x ^{2} + y ^{2} )} \frac{\partial}{\partial x} (a (x^{2} + y^{2}))

\frac{\partial}{\partial x} (a (x^{2} + y^{2})) = 2 a x

= a \frac{1}{a ( x ^{2} + y ^{2} )} \cdot 2 a x

Simplify

a \frac{1}{a ( x ^{2} + y ^{2} )} \cdot 2 a x : \frac{2 a x}{x ^{2} + y ^{2}}

= \frac{2 a x}{x ^{2} + y ^{2}}

\frac{\partial}{\partial x} (x^{3} + y^{3} + z^{3} + 6 x yz)

\frac{2}{s ^{3}} + \frac{6}{s ^{2}} - \frac{3}{s}

y 1 - x^{2} y^{^{'}} - x 1 - y^{2} = 0

\frac{d P}{d t} = P (9 - P) - \frac{81}{4}

\int \frac{3 - 4 x ^{3} + 6 x ^{6}}{x ^{6}} d x

What is (\partial)/(\partial x)(aln(a(x^2+y^2))) ?
The answer to (\partial)/(\partial x)(aln(a(x^2+y^2))) is (2ax)/(x^2+y^2)