Math question. please help.

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1

<p>Ok i am having the hardest time arriving at the answer.</p>

<p>This is really an issue of simpliciation rather than actual calculus </p>

<p>Here is the problem:</p>

<p>Find the derivative:</p>

<p>y=ln(x^4sin(x)^2) Please note its not sin x^2 its (sin x)^2</p>

<p>y'=ln (x^4) +ln (sinx)^2</p>

<p>y'=4x^3/x^4 + 2sinx*cosx/(sinx)^2</p>

<p>y'= 4/x + 2/tanx</p>

<p>y'=4tanx+2x/ (x*tanx)</p>

<p>from this point on , i need to simplify to arrive at </p>

<p>2(x<em>cosx+2sinx)/x</em>sinx</p>

<p>I cant arrive at that dam answer, someone please please hlep? </p>

<p>( i know i am probably missing something stupid)</p>

2

<p>ln(x^4sin^2(x))=4sin^2(x)*ln(x)</p>

3

<p>Or is it ln((x^4)(sin^2(x))? In this case,</p>

<p>y = ln(x^4) + ln(sin^2(x))
y = 4ln(x) + 2ln(sin(x))
y' = 4(1/x) + 2(1/sin(x))(cos(x))
y' = (4sin(x))/(xsin(x)) + (2xcos(x))/(xsin(x))
y' = (2(xcos(x) + 2sin(x)))/(xsin(x))</p>

<p>Reducing to tan's kind of messed you up.</p>