Question

How do you find the derivative of `y=ln(70x^2+24x+7)`?

Answer 1

`y'=(140x+24)/(70x^2+24x+7)`

Explanation:

Note that

`(ln(x))'=1/x`
We also Need the chain rule. So we get

`y'=(140x+24)/(70x^2+24x+7)`

Answer 2

`f'(y)=(140x+24)/(70x^2+24x+7)`

Explanation:

Bit old fashioned in my approach so I will be using format type `dy/dx`

Set `f(x)=t=70x^2+24x+7 color(white)("d") ->(dt)/(dx) = 140x+24`

Set `y=ln(t) ->dy/(dt)=1/t`

Combining these as: `dy/(dt)xx(dt)/dx = dy/dx`

So applying the above:

`dy/dx=f'(x)= color(white)("d")1/(70x^2+24x+7)xx(140x+24)`

`color(white)("dddd.ddddd") = color(white)("d") (140x+24)/(70x^2+24x+7)`