How do you do these problems?

Question

(1). y=(tan^-1)(sqrt(5x^2-1))

dy/dx=??

(2). f(x)=8xarcsin(x)
f '(x)=????

Answer 1

1) 1 / [10x^2 * (5x^2 - 1)^0.5] {or, if one prefers no fractional powers in the denominator: (5x^2 - 1)^0.5 / (50x^4 - 10x^2) }

2) 8x / (1 - x^2)^0.5 + 8 * arcsin (x) {or, if one prefers no fractional powers in the denominator: 8*x*(1 - x^2)^(1/2) / (1 - x^2) + 8*arcsin (x) }


1) I differ from the 2nd answerer. I learned f' ( arctan (m^n) ) = n / (m^2n + 1) * [m^(n-1)]. Since I would set up the problem as m = 5x^2 - 1 and n = 1/2, the derivative would flesh out to be: f'(arctan (5x^2 - 1)^(1/2) = { (1/2) / [(5x^2 - 1)^[2*(1/2)] +1] } * (5x^2 - 1)^(1/2 - 1) and that reduces as follows:

= { (1/2) / [(5x^2 - 1)^[2*(1/2)] +1] } * (5x^2 - 1)^(1/2 - 1) calculate out the exponents
= { (1/2) / [(5x^2 - 1)^1 +1] } * (5x^2 - 1)^(–1/2) remove superfluous exponents
= { (1/2) / [(5x^2 - 1) +1] } * (5x^2 - 1)^(–1/2) combine the "-1) + 1]" terms; remove the negative exponent by taking the term into the denominator and do the same with the divisor fraction
= 1 / { 2 * (5x^2) * (5x^2 - 1)^(1/2).

(Then, as noted in the beginning, if one prefers no fractional powers in the denominator (akin to not leaving, say, √2 in the denominator), then multiply by the fraction: (5x^2 - 1)^(1/2) / (5x^2 - 1)^(1/2) and reduce the denominator.)

I believe the 2nd answerer left out the " * m^(n-1) term.


2) f'(k*m*arcsin (m^n)) = k*(m^n)*n / [1 - m^(2*n)]^(1/2) + k * arcsin (m^n). If we let m = x, n = 1, and k = 8, then the problem becomes f'[8*x*arcsin (x)] = 8*(x^1)*1 / [1 - x^(2*1)]^(1/2) + 8*arcsin (x^1). This reduces as follows:

= 8*(x^1)*1 / [1 - x^(2*1)]^(1/2) + 8*arcsin (x^1) remove the superfluous "^1's," multiply out the instances of "*1"
= 8*x / (1 - x^2)^(1/2) + 8*arcsin (x)

(If one prefers, one can remove the fractional power from the deminator by multiplying the first term by (1 - x^2)^(1/2) / (1 - x^2)^(1/2): = 8*x*(1 - x^2)^(1/2) / (1 - x^2) + 8*arcsin (x).)

Answer 2

y=(tan^-1)(sqrt(5x^2-1))
Let u = (5x^2-1)^.5
du/dx = 10x/(5x^2-1)^.5
y = arctan (u)
y' = 1/(1+u^2) du/dx
y' = 10x/[(5x^2)(5x^2-1)^.5]

f(x)=8xarcsin(x)
y' = 8arcsin(x) + 8x(1/(1-x^2)^.5)

Answer 3

u = (tan^-1)
v = (sqrt(5x^2-1))

Product rule