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

The derivative of sin(ln((x^2+4^4))) is (8xcos(4ln(x^2+4)))/(x^2+4)

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

The first derivative of sin(ln((x^2+4^4))) is (8xcos(4ln(x^2+4)))/(x^2+4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\frac{8 x cos ( 4 ln ( x ^{2} + 4 ) )}{x ^{2} + 4}

Solution steps

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

Apply the chain rule:

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

= cos (ln ((x^{2} + 4)^{4})) \frac{d}{d x} (ln ((x^{2} + 4)^{4}))

\frac{d}{d x} (ln ((x^{2} + 4)^{4})) = \frac{8 x}{x ^{2} + 4}

= cos (ln ((x^{2} + 4)^{4})) \frac{8 x}{x ^{2} + 4}

Simplify

cos (ln ((x^{2} + 4)^{4})) \frac{8 x}{x ^{2} + 4} : \frac{8 x cos ( 4 ln ( x ^{2} + 4 ) )}{x ^{2} + 4}

= \frac{8 x cos ( 4 ln ( x ^{2} + 4 ) )}{x ^{2} + 4}

7 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0

\int \frac{2 x + 2}{x ^{2} - 2 x + 1} d x

x \to 8 lim (\frac{x ^{2} - 5}{x - 8})

\int x (3 x^{2} + 1) d x

\frac{d y}{d x} = \frac{e ^{x}}{y}

What is the derivative of sin(ln((x^2+4^4))) ?
The derivative of sin(ln((x^2+4^4))) is (8xcos(4ln(x^2+4)))/(x^2+4)
What is the first derivative of sin(ln((x^2+4^4))) ?
The first derivative of sin(ln((x^2+4^4))) is (8xcos(4ln(x^2+4)))/(x^2+4)