Derivatives: Logarithmic Differentiation

So i solved a few of the steps but i'm stuck at this one step and can't move on :(

Find yhttps://lewis.chem.sfu.ca/adm/jsMath.../85/char30.png if

y=(10x+9)^10x+1https://lewis.chem.sfu.ca/adm/jsMath...120/char3A.png

Note: The product of two numbers or two expressions should be entered as a*b. For the quotient of two numbers use a/b. For the power use a^b.

1. Take the natural logarithm of both sides:
lny = ln (10*x+9)^(10*x+1)

2. Simplify by using properties of logarithms:
Important! In your answer, lna must be entered as log(a).
lny = (10*x+1)*log(10*x+9)

3. Differentiate both sides with respect to x:
yhttps://lewis.chem.sfu.ca/adm/jsMath.../85/char30.png https://lewis.chem.sfu.ca/adm/jsMath...120/char3D.pngy = (Help please)

4. Solve for yhttps://lewis.chem.sfu.ca/adm/jsMath.../85/char30.png:
yhttps://lewis.chem.sfu.ca/adm/jsMath.../85/char30.png = yhttps://lewis.chem.sfu.ca/adm/jsMath...120/char01.png( )

5. Substitute the original expression for y to get yhttps://lewis.chem.sfu.ca/adm/jsMath.../85/char30.png in terms of x only:
yhttps://lewis.chem.sfu.ca/adm/jsMath.../85/char30.png =

If someone could help me from 3 onwards i would really appreciate it.

Thanks :)

can't you use $\frac{d(uv)}{dx}=\frac{du}{dx}.v+\frac{dv}{dx}.u$

then $y'y =(10x+1) \frac{10}{(10x+9)} + 10ln(10x+9)$

Ah that was it, thanks a lot!