What is the derivative of ln(sin(4x^3+6x^2+8)) ?

The derivative of ln(sin(4x^3+6x^2+8)) is cot(4x^3+6x^2+8)(12x^2+12x)

What is the first derivative of ln(sin(4x^3+6x^2+8)) ?

The first derivative of ln(sin(4x^3+6x^2+8)) is cot(4x^3+6x^2+8)(12x^2+12x)

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\frac{d}{d x} (ln (sin (4 x^{3} + 6 x^{2} + 8)))

cot (4 x^{3} + 6 x^{2} + 8) (12 x^{2} + 12 x)

Solution steps

\frac{d}{d x} (ln (sin (4 x^{3} + 6 x^{2} + 8)))

Apply the chain rule:

\frac{1}{sin ( 4 x ^{3} + 6 x ^{2} + 8 )} \frac{d}{d x} (sin (4 x^{3} + 6 x^{2} + 8))

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

\frac{d}{d x} (sin (4 x^{3} + 6 x^{2} + 8)) = cos (4 x^{3} + 6 x^{2} + 8) (12 x^{2} + 12 x)

= \frac{1}{sin ( 4 x ^{3} + 6 x ^{2} + 8 )} cos (4 x^{3} + 6 x^{2} + 8) (12 x^{2} + 12 x)

Simplify

\frac{1}{sin ( 4 x ^{3} + 6 x ^{2} + 8 )} cos (4 x^{3} + 6 x^{2} + 8) (12 x^{2} + 12 x) : cot (4 x^{3} + 6 x^{2} + 8) (12 x^{2} + 12 x)

= cot (4 x^{3} + 6 x^{2} + 8) (12 x^{2} + 12 x)

\int \frac{2 x + 5}{x ^{2}} d x

\int \frac{1}{( x ^{2} + 4 x + 5 )} d x

\frac{d}{d x} (4 - \frac{600}{x ^{2}})

y = 2 x^{2}, (- 1, 2)

\frac{\partial}{\partial x} (- 2 e^{x})

What is the derivative of ln(sin(4x^3+6x^2+8)) ?
The derivative of ln(sin(4x^3+6x^2+8)) is cot(4x^3+6x^2+8)(12x^2+12x)
What is the first derivative of ln(sin(4x^3+6x^2+8)) ?
The first derivative of ln(sin(4x^3+6x^2+8)) is cot(4x^3+6x^2+8)(12x^2+12x)