What is the derivative of y=ln(3x^2+7y^2) ?

The derivative of y=ln(3x^2+7y^2) is (6x)/(3x^2+7y^2)

What is the first derivative of y=ln(3x^2+7y^2) ?

The first derivative of y=ln(3x^2+7y^2) is (6x)/(3x^2+7y^2)

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derivative of

y = ln (3 x^{2} + 7 y^{2})

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

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\frac{d}{d x} (0.02 x^{3} - 0.07 x^{2} + 8 x)

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\frac{d}{d x} (\frac{4 - 3 x - x ^{2}}{x ^{2} - 1})

What is the derivative of y=ln(3x^2+7y^2) ?
The derivative of y=ln(3x^2+7y^2) is (6x)/(3x^2+7y^2)
What is the first derivative of y=ln(3x^2+7y^2) ?
The first derivative of y=ln(3x^2+7y^2) is (6x)/(3x^2+7y^2)