What is the derivative of f(x)=ln(6-x^2+2x^4) ?

The derivative of f(x)=ln(6-x^2+2x^4) is (8x^3-2x)/(6-x^2+2x^4)

What is the first derivative of f(x)=ln(6-x^2+2x^4) ?

The first derivative of f(x)=ln(6-x^2+2x^4) is (8x^3-2x)/(6-x^2+2x^4)

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derivative of

f (x) = ln (6 - x^{2} + 2 x^{4})

\frac{8 x ^{3} - 2 x}{6 - x ^{2} + 2 x ^{4}}

Solution steps

\frac{d}{d x} (ln (6 - x^{2} + 2 x^{4}))

Apply the chain rule:

\frac{1}{6 - x ^{2} + 2 x ^{4}} \frac{d}{d x} (6 - x^{2} + 2 x^{4})

= \frac{1}{6 - x ^{2} + 2 x ^{4}} \frac{d}{d x} (6 - x^{2} + 2 x^{4})

\frac{d}{d x} (6 - x^{2} + 2 x^{4}) = 8 x^{3} - 2 x

= \frac{1}{6 - x ^{2} + 2 x ^{4}} (8 x^{3} - 2 x)

Simplify

\frac{1}{6 - x ^{2} + 2 x ^{4}} (8 x^{3} - 2 x) : \frac{8 x ^{3} - 2 x}{6 - x ^{2} + 2 x ^{4}}

= \frac{8 x ^{3} - 2 x}{6 - x ^{2} + 2 x ^{4}}

\int t^{3} (1 + t^{4})^{3} d t

\frac{d}{d x} (2 (x - 3))

\frac{d y}{d x} = \frac{( x ^{2} y - 32 )}{16 - x ^{2}} + 2

f (x) = \frac{x ^{2}}{x + 5}, at x = 3

What is the derivative of f(x)=ln(6-x^2+2x^4) ?
The derivative of f(x)=ln(6-x^2+2x^4) is (8x^3-2x)/(6-x^2+2x^4)
What is the first derivative of f(x)=ln(6-x^2+2x^4) ?
The first derivative of f(x)=ln(6-x^2+2x^4) is (8x^3-2x)/(6-x^2+2x^4)