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..
