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)#