How do you differentiate ${log}_{3} (9 x^{2} sin (9 x^{2}))$ $log_3(9x^2sin(9x^2) )$ ?

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Answer 1 · 2017-03-31T21:02:05

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Use the following Properties of Logarithm to expand the problem before taking derivatives.

${log}_{b} (x y) = {log}_{b} x + {log}_{b} y$ $color(red)(log_b(xy)=log_bx+log_by$
${log}_{b} (\frac{x}{y}) = {log}_{b} x - {log}_{b} y$ $color(red)(log_b(x/y)=log_bx-log_by$
${log}_{b} x^{n} = n {log}_{b} x$ $color(red)(log_b x^n =n log_bx$

Then use the formula $\frac{d}{d x} ({log}_{b} x) = \frac{1}{x ln b}$ $color(red)(d/dx(log_bx)=1/(xln b)$ to find the derivative

$y = {log}_{3} (9 x^{2} sin (9 x^{2}))$ $y=log_3(9x^2sin(9x^2))$

$y = {log}_{3} 9 + {log}_{3} x^{2} + {log}_{3} sin (9 x^{2})$ $y=log_3 9 + log_3x^2+log_3 sin(9x^2)$

$y = {log}_{3} 9 + 2 {log}_{3} x + {log}_{3} sin (9 x^{2})$ $y=log_3 9 + 2 log_3x+log_3 sin(9x^2)$

$color(blue)(y'=0+2/(xln3)+1/(sin(9x^2)ln3) * cos (9x^2)*18x$

$color(blue)(y'=2/(xln3)+(cos (9x^2)*18x)/(sin(9x^2)ln3)$

$color(blue)(y'=2/(xln3)+((18x) cot(9x^2))/ln3$

$color(blue)(y'=((18x^2) cot(9x^2))/(xln3)$

How do you differentiate log_3(9x^2sin(9x^2) ) log3(9x2sin(9x2))log_3(9x^2sin(9x^2) ) ?