Question

How can I solve this using Log differentiation? I'm not sure how to even start it.
y=(cos(x)(x^2+9)^(2/3)/(sqr(x^3+2))

Answer 1

y=(cos(x)(x^2+9)^(2/3)/(sqr(x^3+2))
take ln of both sides
ln y=ln ( (cos(x)(x^2+9)^(2/3)/(sqr(x^3+2)) )
by the rules of logs
ln y = ln cosx + ln (x^2 + 9)^(2/3) - ln (x^3+2)^(1/2)
= ln cosx + (2/3)ln (x^2 + 9) - (1/2)ln (x^3 + 2)

now differentiate ...

y' /y = sinx/cosx + (2/3)(2x)/(x^2 + 9) - (1/2)(3x^2)/(x^3 + 2)
= tanx + (2/3)(2x)/(x^2 + 9) - (1/2)(3x^2)/(x^3 + 2)

y ' = (cos(x)(x^2+9)^(2/3)/(sqr(x^3+2))( tanx + (2/3)(2x)/(x^2 + 9) - (1/2)(3x^2)/(x^3 + 2) )

check this, especially my typing, so easy to make mistakes in this mess