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

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Answer 1 · 2018-06-25T12:06:00

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

Note that

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

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

Answer 2 · 2018-06-25T12:30:46

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

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:

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

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

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