find the first and second derivative of this function. (please explain work)?

Question

Function 1 = a/(s^2 +a^2) (a is a constant)

Function 2 = s/(s^2 + a^2) (a is a constant)

Answer 1

Function 1 = a/(s^2 +a^2) (a is a constant)

Function 2 = s/(s^2 + a^2) (a is a constant)


Ley y = a/(s^2 +a^2)

dy/ds = d(a/(s^2 +a^2))/ds
=d/ds(a*(s^2 +a^2)^-1)
=a*d/ds((s^2 +a^2)^-1) ..........[d/dx(af(x))=ad/dx(f(x))
=a*(-1)*(s^2 +a^2)^(-2)*d/dS(s^2 +a^2).........[d/dx(f(x)^n)=n*f(x)^(n-1)*d/dx(f(x))
=-a(s^2 +a^2)^(-2)*d/ds(s^2)..........[d/dx(a)=0 where a is a constant
=-a(s^2 +a^2)^(-2)*(2)s
=-2as/(s^2 +a^2)^2
.......(1)

Ley y = s/(s^2 + a^2)
dy/ds = d(s/(s^2 + a^2))/ds
=d/ds(s*(s^2 + a^2)^(-1))
=s*d/ds((s^2 + a^2)^(-1))+(s^2 + a^2)^(-1)*d/ds(s) .....[d/dx(f1(x)*f2(x))=f1(x)*d/dx(f2(x))+f2(x)*d/dx(f1(x))
=s*(-2s/(s^2 +a^2)^2)+s*(s^2 + a^2)^(-1)...[From (1)
=-2s^2/(s^2 + a^2)^(2)-1/(s^2 + a^2)
=-(2s^2+s^2 + a^2)/(s^2 + a^2)^(2)
=-(3s^2 + a^2)/(s^2 + a^2)^(2)

Answer 2

1) f' = 2ax / (x^2 + a^2)^2
f'' = 2a / (x^2 + a^2)^2 - 8ax^2/(x^2 + a^2)^3

i use the rule : f(x) = g(h(x)) , f' = g'(h(x)) * h'(x)