How do you differentiate $y = {log}_{2} (x^{4} sin x)$ $y = log_2 (x^4sinx)$ ?

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Answer 1 · 2016-06-28T18:23:21

$y' = (4 sin x + x cos x)/(ln 2 * xsinx)$

$y = log_2 (x^4sinx)$

first switch into natural logs as these are calculus friendly

$y = (ln (x^4sinx))/ln 2 = implies ln 2 \ y = ln (x^4sinx)$

so therefore by the chain rule

$ln2 \ y' = 1/(x^4sinx)*(x^4sinx)'$

$(x^4sinx)' = (x^4)' sin x + x^4 (sin x)'$
$= 4x^3 sin x + x^4 cos x$
$= x^3(4 sin x + x cos x)$

so the derivative is
$ln2 \ y' = 1/(x^4sinx)*x^3(4 sin x + x cos x)$
$= (4 sin x + x cos x)/(xsinx)$

so

$y' = (4 sin x + x cos x)/(ln 2 * xsinx)$

How do you differentiate y = log_2 (x^4sinx)y=log2(x4sinx)y = log_2 (x^4sinx)?