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- 10 years agoFavourite answer
y'=(1/(3x tanx)(ln6))(3 tanx + 3x sec^2 x)
y'=(3(tanx + x sec^2 x) / (3(x tanx ln6))
y'=(tanx+x sec^2 x)/(x tanx ln6)
because:
d/dx(log(b)x)=(1/x lnb)*d/dx(x)
d/dx(ab)=d/dx(a)b+d/dx(b)a
d/dx(tanx)=sec^2 x
Source(s): Calculus 1 currently enrolled - 10 years ago
y' = log 6 ( (3x) (secxtanx) + (3x tanx) (3)) + (3x tanx) ( 10 ln 6) < product rule within the product rule inside. >
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