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

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Answer 1 · 2016-06-26T19:57:09

$= \frac{2}{ln 2} (\frac{1}{x} + x {cot x}^{2})$ $= 2/(ln 2) ( 1/x + x cot x^2)$

$(log_2(x^2sinx^2))'$

first we need to shift the base into natural logs as these are just made for calculus

so using $log_a b = (log_c b)/(log_c a)$

so
$log_2(x^2sinx^2) = (ln (x^2sinx^2))/(ln 2) = 1/(ln 2)ln (x^2sinx^2)$

then we use another standard result namely that

$(ln f(x))' = 1/(f(x))*f'(x)$

so $( 1/(ln 2)ln (x^2sinx^2))' = 1/(ln 2)( ln (x^2sinx^2))'$

$= 1/(ln 2)1/ (x^2sinx^2) (x^2sinx^2)' \qquad square$

for $(x^2sinx^2)'$ we use the product rule

$(x^2sinx^2)' = (x^2)'sinx^2 + x^2 (sinx^2)'$

$= 2x sinx^2 + x^2 (sinx^2)' \qquad star$

for $(sinx^2)'$ we use the chain rule so

$(sinx^2)' = cos x^2 (2x) = 2x cos x^2$

pop that back into $star$ to get

$(x^2sinx^2)' = 2x sinx^2 + x^2 * 2x cos x^2$
$\implies (x^2sinx^2)' = 2x sinx^2 + 2x^3 cos x^2$

pop that back into $square$ for

$( 1/(ln 2)ln (x^2sinx^2))' = 1/(ln 2)1/ (x^2sinx^2) (x^2sinx^2)'$

$= 1/(ln 2)1/ (x^2sinx^2) (2x sinx^2 + 2x^3 cos x^2)$

this can be simplified in any number of ways, so i'm going for brevity

$= 2/(ln 2) ( 1/x + x cot x^2)$

How do you differentiate log_2(x^2sinx^2) log2(x2sinx2)log_2(x^2sinx^2) ?