What is (\partial)/(\partial x)(ln(4x^2+9y^2+8)) ?

The answer to (\partial)/(\partial x)(ln(4x^2+9y^2+8)) is (8x)/(4x^2+9y^2+8)

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(\partial)/(\partial x)(ln(4x^2+9y^2+8))

\frac{\partial}{\partial x} (ln (4 x^{2} + 9 y^{2} + 8))

\frac{8 x}{4 x ^{2} + 9 y ^{2} + 8}

Solution steps

\frac{\partial}{\partial x} (ln (4 x^{2} + 9 y^{2} + 8))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{4 x ^{2} + 9 y ^{2} + 8} \frac{\partial}{\partial x} (4 x^{2} + 9 y^{2} + 8)

= \frac{1}{4 x ^{2} + 9 y ^{2} + 8} \frac{\partial}{\partial x} (4 x^{2} + 9 y^{2} + 8)

\frac{\partial}{\partial x} (4 x^{2} + 9 y^{2} + 8) = 8 x

= \frac{1}{4 x ^{2} + 9 y ^{2} + 8} \cdot 8 x

Simplify

\frac{1}{4 x ^{2} + 9 y ^{2} + 8} \cdot 8 x : \frac{8 x}{4 x ^{2} + 9 y ^{2} + 8}

= \frac{8 x}{4 x ^{2} + 9 y ^{2} + 8}

y^{^{''}} + 2 y = 1

y^{^{''}} + 2 y = sin (x 2)

t an g e n t f (x) = 4 x^{2} + 12 x + 7, at x = - 2

d er i v a t i v e x^{2} - x

t a y l or x^{- 1}

What is (\partial)/(\partial x)(ln(4x^2+9y^2+8)) ?
The answer to (\partial)/(\partial x)(ln(4x^2+9y^2+8)) is (8x)/(4x^2+9y^2+8)