How would you differentiate ln(x^2+3x+5)?
Here we need to use the chain rule because we have a function (natural log) of another function (x^2+3x+5). Let u=x^2+3x+5, and differentiate lnu with respect to u, this gives us 1/u. Then we differentiate x^2+3x+5 with respect to x, so we get 2x+3. Now the chain rule says: dy/dx=dy/dudu/dx, so we have dy/dx = (1/u)(2x+3)=(2x+3)/(x^2+3x+5)
OH
Related Maths A Level answers
All answers ▸Find the tangent to the curve y = x^2 + 3x + 2 at x = 1
Find the area encompassed by y=(3-x)x^2 and y=x(4-x) between x=0 and x=2.
Find dy/dx in terms of t of the parametric equations x=4e^-2t, y=4 - 2e^2t
What is the area bound by the x-axis, the lines x=1 and x=3 and the curve y=3x^(2)-1/x ? Answer in exact form.