Step 1 : we have #y=(2x^2-1)/(xsqrt(1+x^2))#
This equation is of the form #f(x)=g(x)/(h(x))#
Where, #g(x)=2x^2-1# and #h(x)=xsqrt(1+x^2)#
We will use the quotient rule to find the derivative.
#=>f'(x)=(g'(x)h(x)-h'(x)g(x))/(h^2(x))#
Step 2 : Lets solve the different parts:
#g(x)=2x^2-1#
#=>color(red)(g'(x)=4x)#
Step 3 : Next, to solve #h'(x)#
We have #h(x)=xsqrt(1+x^2)#
This is of the form #h(x)=a(x)*b(x)#
where, #a(x)=x# and #b(x)=sqrt (1+x^2)#
Apply product rule:
#=>color(brown)(h'(x)=a'(x)b(x)+b'(x)a(x))#
#a(x)=x#
#=>color(blue)(a'(x)=1)#
#b(x)=sqrt (1+x^2)#
#=>b(x)=(1+x^2)^(1/2)#
#=>b'(x)=1/2(1+x^2)^(1/2-1)xx2x#
Use #color(green)(f(x)=x^n=>f'(x)=nx^(n-1))#
#=>b'(x)=(1+x^2)^(-1/2)xxx#
#=>b'(x)=1/(1+x^2)^(1/2)xxx#
#=>color(blue)(b'(x)=1/sqrt(1+x^2)xxx)#
back to #color(brown)(h'(x)=a'(x)b(x)+b'(x)a(x))#
Put back the values to find #h'(x)#
#=>h'(x)=sqrt (1+x^2)+1/sqrt(1+x^2)xx x xx x#
#=>h'(x)=(1+x^2+x^2)/sqrt(1+x^2)#
#=>color(red)(h'(x)=(1+2x^2)/sqrt(1+x^2))#
Final step : Put back the values in #f'(x)# and simplify:
#f'(x)=(g'(x)h(x)-h'(x)g(x))/(h^2(x))#
#=>f'(x)=(4x xxxsqrt(1+x^2) - (1+2x^2)/sqrt(1+x^2) xx (2x^2-1))/(xsqrt(1+x^2))^2 #
#=>f'(x)=((4x^2 xx (1+x^2) - (1+2x^2) xx (2x^2-1))/(sqrt(1+x^2)))/((x^2)(sqrt(1+x^2))^2) #
#=>f'(x)=(4x^2 xx (1+x^2) - (1+2x^2) xx (2x^2-1))/((x^2)(sqrt(1+x^2))^3) #
#=>f'(x)=((4x^2+4x^4) - (4x^4-1))/((x^2)(sqrt(1+x^2))^3) #
#=>f'(x)=(4x^2+4x^4 -4x^4+1)/((x^2)(sqrt(1+x^2))^3) #
#=>f'(x)=(4x^2+1)/((x^2)(sqrt(1+x^2))^3) #
This looks long and confusing. But if you solve one step at a time, its a piece of cake.
All the best..
