Differentiate with respect to x: y = ln(x^2+4*x+2).

Let u = x2+4x+2 so y = ln(u).

Then dy/du = 1/u and du/dx = 2x+4.

Using the chain rule we have:

dy/dx = (dy/du)*(du/dx)

= (1/u)*(2x+4)

= (2x+4)/(x2+4x+2).

OL
Answered by Okim L. Maths tutor

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make into a cartesian equation= x=ln(t+3) y= 1/t+5


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