Answer:
h'(x) = (-x² ln x + x² + 1) / (x (x² + 1)^(³/₂))
Step-by-step explanation:
h(x) = ln x / √(x² + 1)
You can either use quotient rule, or you can rewrite using negative exponents and use product rule.
h(x) = (ln x) (x² + 1)^(-½)
h'(x) = (ln x) (-½) (x² + 1)^(-³/₂) (2x) + (1/x) (x² + 1)^(-½)
h'(x) = (-x ln x) (x² + 1)^(-³/₂) + (1/x) (x² + 1)^(-½)
h'(x) = (x² + 1)^(-³/₂) (-x ln x + (1/x) (x² + 1))
h'(x) = (1/x) (x² + 1)^(-³/₂) (-x² ln x + x² + 1)
h'(x) = (-x² ln x + x² + 1) / (x (x² + 1)^(³/₂))
Answer:
9
Step-by-step explanation:
Answer:
#1. Identity #2. 0 #3. No solution
Step-by-step explanation:
#1.
5y + 2 = (1/2)(10y+4)
5y + 2 = 5y + 2
This would be identity as the equation of the left and right are the same. This is not to be confused with no solution(explained below).
#2.
0.5b + 4 = 2(b+2)
0.5b + 4 = 2b + 4
0.5 b - 2b = 0
b = 0
#3.
-3x + 5 = -3x + 10
This equation has no solution because when you try to bring the -3x to one side, the x variable itself gets eliminated. So, how is it different from identity? Well in the first equation, it is true that when we try to bring the 5y one side it eliminates the y variable, however, that is also true for the constants(since if we try to bring the 2 to one side, it will be 2-2 which will equal 0, thus eliminating each other), but in this case, even if we remove the x, the constants will not equal 0, thus it will have no solution.
Answer:
69a-10
Step-by-step explanation:
Answer:
2132c1 23
Step-by-step explanation:
a123231
