What is (\partial)/(\partial x)(ln(6x^2+7y^2+3)) ?

The answer to (\partial)/(\partial x)(ln(6x^2+7y^2+3)) is (12x)/(6x^2+7y^2+3)

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(\partial)/(\partial x)(ln(6x^2+7y^2+3))

\frac{\partial}{\partial x} (ln (6 x^{2} + 7 y^{2} + 3))

\frac{12 x}{6 x ^{2} + 7 y ^{2} + 3}

Solution steps

\frac{\partial}{\partial x} (ln (6 x^{2} + 7 y^{2} + 3))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{6 x ^{2} + 7 y ^{2} + 3} \frac{\partial}{\partial x} (6 x^{2} + 7 y^{2} + 3)

= \frac{1}{6 x ^{2} + 7 y ^{2} + 3} \frac{\partial}{\partial x} (6 x^{2} + 7 y^{2} + 3)

\frac{\partial}{\partial x} (6 x^{2} + 7 y^{2} + 3) = 12 x

= \frac{1}{6 x ^{2} + 7 y ^{2} + 3} \cdot 12 x

Simplify

\frac{1}{6 x ^{2} + 7 y ^{2} + 3} \cdot 12 x : \frac{12 x}{6 x ^{2} + 7 y ^{2} + 3}

= \frac{12 x}{6 x ^{2} + 7 y ^{2} + 3}

\frac{\partial}{\partial x} (x^{4} y^{6} + 4 x^{7} y)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 3 e^{x}

t^{2} e^{2 t}

\int_{- 7}^{7} x 49 - x^{2} d x

d y - d x + y = e^{4 x}

What is (\partial)/(\partial x)(ln(6x^2+7y^2+3)) ?
The answer to (\partial)/(\partial x)(ln(6x^2+7y^2+3)) is (12x)/(6x^2+7y^2+3)