How do you differentiate $y = cos ln 4 x^{3}$ $y=cosln4x^3$ ?

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Answer 1 · 2016-10-31T17:44:23

$\frac{d y}{d x} = \frac{- 3 sin (ln (4 x^{3}))}{x}$ $dy/dx=(-3sin(ln(4x^3)))/x$

$y = cos ln 4 x^{3}$ $y=cosln4x^3$

Use chain rule $(f(g(x))' = f'(g(x)) * g'(x)$

$dy/dx=-sin(ln(4x^3))*1/(4x^3) *12x^2$

$dy/dx=-sin(ln(4x^3)) *3/x$

$dy/dx=(-3sin(ln(4x^3)))/x$

How do you differentiate y=cosln4x^3y=cosln4x3y=cosln4x^3?