How do you differentiate $y=lnx(x^5+10x^2-19)$ ?

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Answer 1 · 2016-08-07T01:31:24

$(dy)/(dx)=1/x(x^5+10x^2-19)+5x(x^3+4)lnx$

Using product rule, if $y=lnx(x^5+10x^2-19)$

$(dy)/(dx)=1/x(x^5+10x^2-19)+lnx(5x^4+20x)$

= $1/x(x^5+10x^2-19)+5x(x^3+4)lnx$

How do you differentiate y=lnx(x^5+10x^2-19)y=lnx(x^5+10x^2-19)?