How do you differentiate $y=3^sinx$ ?

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Answer 1 · 2016-09-24T17:05:28

$y = 3^sinx$

$lny = ln(3^(sinx))$

$lny = sinxln3$

$1/y(dy/dx) = (cosx xx ln3 + 0 xx sinx)$

$dy/dx = (ln3cosx)/(1/y)$

$dy/dx = y xx ln3cosx$

$dy/dx = 3^sinx xx ln3cosx$

$dy/dx = 3^sinxln3cosx$

Hopefully this helps!

How do you differentiate y=3^sinxy=3^sinx?