What is (\partial)/(\partial x)(ln(7x^2+y^2+6)) ?

The answer to (\partial)/(\partial x)(ln(7x^2+y^2+6)) is (14x)/(7x^2+y^2+6)

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(\partial)/(\partial x)(ln(7x^2+y^2+6))

\frac{\partial}{\partial x} (ln (7 x^{2} + y^{2} + 6))

\frac{14 x}{7 x ^{2} + y ^{2} + 6}

Solution steps

\frac{\partial}{\partial x} (ln (7 x^{2} + y^{2} + 6))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{7 x ^{2} + y ^{2} + 6} \frac{\partial}{\partial x} (7 x^{2} + y^{2} + 6)

= \frac{1}{7 x ^{2} + y ^{2} + 6} \frac{\partial}{\partial x} (7 x^{2} + y^{2} + 6)

\frac{\partial}{\partial x} (7 x^{2} + y^{2} + 6) = 14 x

= \frac{1}{7 x ^{2} + y ^{2} + 6} \cdot 14 x

Simplify

\frac{1}{7 x ^{2} + y ^{2} + 6} \cdot 14 x : \frac{14 x}{7 x ^{2} + y ^{2} + 6}

= \frac{14 x}{7 x ^{2} + y ^{2} + 6}

8 x^{3}

x y^{^{'}} + 2 y = - 2 e^{x}

x \to - 3 lim (\frac{7}{x + 3})

y^{^{'}} = t^{2} - y

\frac{d}{d x} (arccos (\frac{3}{x}))

What is (\partial)/(\partial x)(ln(7x^2+y^2+6)) ?
The answer to (\partial)/(\partial x)(ln(7x^2+y^2+6)) is (14x)/(7x^2+y^2+6)