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

2 Answers
Jun 25, 2018

#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)#

Jun 25, 2018

#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)#