How do you find the derivative of $y = ln[(x+3)/x^2)]$ ?

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Answer 1 · 2016-12-02T21:01:08

$dy/dx = -(x +6)/(x(x +3))$

We let $y = lnu$ and $u = (x + 3)/x^2$

Then $dy/(du) = 1/u$ and $(du)/dx = (1(x^2) - 2x(x + 3))/(x^2)^2 = (x^2 - 2x^2 - 6x)/x^4 = (-x(x + 6))/x^4 = -(x + 6)/x^3$

The chain rule states that $color(red)(dy/dx= dy/(du) xx (du)/dx$ .

$dy/dx = 1/u xx -(x + 6)/x^3$

$dy/dx= (-(x + 6)/x^3)/((x + 3)/x^2)$

$dy/dx= -(x + 6)/x^3 xx x^2/(x + 3)$

$dy/dx = -(x +6)/(x(x +3))$

Hopefully this helps!

Answer 2 · 2016-12-02T21:58:02

We can also rewrite this the logarithm rules $log(a/b)=log(a)-log(b)$ and $log(c^d)=dlog(c)$ .

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

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

We should know that $d/dxln(x)=1/x$ . Applying the chain rule to this shows us that $d/dxln(u)=1/u*(du)/dx$ .

Then:

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

Since $d/dx(x+3)=1$ :

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

Finding a common denominator:

$dy/dx=(x-2(x+3))/(x(x+3))$

$color(white)(dy/dx)=(-x-6)/(x(x+3))$

$color(white)(dy/dx)=-(x+6)/(x(x+3))$

How do you find the derivative of y = ln[(x+3)/x^2)]y = ln[(x+3)/x^2)]?