How do you find the derivative of x/(1+x^2)x1+x2?

1 Answer
Jun 6, 2015

You can do this two ways.

a)
Quotient Rule:
d/(dx)[(f(x))/(g(x))] = (g(x)(df(x))/(dx) - f(x)(dg(x))/(dx))/((g(x))^2ddx[f(x)g(x)]=g(x)df(x)dxf(x)dg(x)dx(g(x))2

d/(dx)[x/(1+x^2)] = [(1+x^2)(1) - (x)(2x)]/(1+x^2)^2ddx[x1+x2]=(1+x2)(1)(x)(2x)(1+x2)2

= (1+x^2-2x^2)/(1+x^2)^2=1+x22x2(1+x2)2

= (1-x^2)/[(1+x^2)^2]=1x2(1+x2)2

b)
Product Rule + Chain Rule:

d/(dx)[f(x)*g(x)] = f(x)(dg(x))/(dx) + g(x)(df(x))/(dx)ddx[f(x)g(x)]=f(x)dg(x)dx+g(x)df(x)dx

d/(dx)[x*(1/(1+x^2))] = d/(dx)[x*(1+x^2)^(-1)]ddx[x(11+x2)]=ddx[x(1+x2)1]

= [x][-(1+x^2)^(-2)*(2x)] + [1/(1+x^2)][1]=[x][(1+x2)2(2x)]+[11+x2][1]

= -(2x^2)/(1+x^2)^2 + 1/(1+x^2)=2x2(1+x2)2+11+x2

= [-2x^2 + 1 + x^2]/(1+x^2)^2=2x2+1+x2(1+x2)2

= [1 - x^2]/(1+x^2)^2=1x2(1+x2)2