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

The answer to (\partial)/(\partial x)(6ln(6x^3+y^2)) is (108x^2)/(6x^3+y^2)

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

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

\frac{108 x ^{2}}{6 x ^{3} + y ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (6 x^{3} + y^{2}) = 18 x^{2}

= 6 \cdot \frac{1}{6 x ^{3} + y ^{2}} \cdot 18 x^{2}

Simplify

6 \cdot \frac{1}{6 x ^{3} + y ^{2}} \cdot 18 x^{2} : \frac{108 x ^{2}}{6 x ^{3} + y ^{2}}

= \frac{108 x ^{2}}{6 x ^{3} + y ^{2}}

g (t) = t^{3} cos (t)

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

x \to - 0.00000001 lim (x^{x})

f (x) = [x + (x + sin^{2} (x))^{3}]^{4}

f (x) = x^{2} + 5 - 3 x^{- 2}

What is (\partial)/(\partial x)(6ln(6x^3+y^2)) ?
The answer to (\partial)/(\partial x)(6ln(6x^3+y^2)) is (108x^2)/(6x^3+y^2)