How do you differentiate $y =-ln [ 3+ (9+x^2) / x]$ ?

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Answer 1 · 2017-12-24T11:17:56

$dy/dx=-1/(3+9/x+x)*(1-9/x^2)$

$y=-ln(3+(9+x^2)/x)=-ln(3+9/x+x)$

Use the chain rule $dy/dx=dy/(du)*(du)/dx$

Let $y=-ln(u)$ then $dy/(du)=-1/u$
And $u=3+9/x+x$ then $(du)/(dx)=-9/x^2+1$

$dy/dx=-1/u*(1-9/x^2)$

Substitute $u=3+9/x+x$

$dy/dx=-1/(3+9/x+x)*(1-9/x^2)$

How do you differentiate y =-ln [ 3+ (9+x^2) / x] y =-ln [ 3+ (9+x^2) / x]?