How do you find the first and second derivative of #ln[lnx^2+1)]#?

Question

Charles Anctil · Accepted Answer

# d/dx ln \ ln(x^2+1) \ = (2x)/((x^2+1)ln(x^2+1)) #
# d^2/(dx^2) ln \ ln(x^2+1) = ( 2((1-x^2)ln(x^2+1) - 2x^2)) / ((x^2+1)^2ln^2(x^2+1)) #

Suppose we let: # y= ln \ ln(x^2+1) # Using the chain rule we have: # dy/dx = 1/(ln(x^2+1)) * d/dx(ln(x^2+1)) # # " " = 1/(ln(x^2+1)) * 1/(x^2+1) * d/dx(x^2+1) # # " " = 1/(ln(x^2+1)) * 1/(x^2+1) * 2x # # " " = (2x)/((x^2+1)ln(x^2+1)) # To find the second derivative we will need to apply the quotient rule: # (d^2y)/(dx^2) = ( ((x^2+1)ln(x^2+1))(d/dx 2x) - (2x)(d/dx (x^2+1)ln(x^2+1)) ) / ((x^2+1)ln(x^2+1))^2 # # " " = ( 2(x^2+1)ln(x^2+1) - (2x)(d/dx (x^2+1)ln(x^2+1)) ) / ((x^2+1)^2ln^2(x^2+1)) # And using the product rule, combined with the chain rulewe have: # d/dx (x^2+1)ln(x^2+1) = (x^2+1)(d/dx ln(x^2+1)) + (d/dx (x^2+1))(ln(x^2+1)) # # " " = (x^2+1)(1/(x^2+1)d/dx (x^2+1)) + (2x)(ln(x^2+1)) # # " " = 2x + 2xln(x^2+1) # Injecting this result into our earlier findings we get: # (d^2y)/(dx^2) = ( 2(x^2+1)ln(x^2+1) - (2x)(2x + 2xln(x^2+1)) ) / ((x^2+1)^2ln^2(x^2+1)) # # " " = ( 2{(x^2+1)ln(x^2+1) - x(2x + 2xln(x^2+1))} ) / ((x^2+1)^2ln^2(x^2+1)) # # " " = ( 2{(x^2+1)ln(x^2+1) - 2x^2 - 2x^2ln(x^2+1)}) / ((x^2+1)^2ln^2(x^2+1)) # # " " = ( 2{(x^2+1-2x^2)ln(x^2+1) - 2x^2}) / ((x^2+1)^2ln^2(x^2+1)) # # " " = ( 2((1-x^2)ln(x^2+1) - 2x^2)) / ((x^2+1)^2ln^2(x^2+1)) #

Emma Aldridge · Answer

undefined To find the first derivative, use the chain rule and the fact that the derivative of ln(u) is u'/u, where u is a function of x. For the second derivative, apply the chain rule and the derivative of 1/u, which is -u'/u^2.