#frac{4x-3}{sqrt{2x^2+1}}#
First of all, we have a quotient. The derivative of a quotient #f(x)/g(x)# is given by #frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}#. So we get that
#d/dx [frac{4x-3}{sqrt{2x^2+1}}]=frac{d/dx[4x-3]sqrt{2x^2+1}-(4x-3)d/dx [sqrt{2x^2+1}]}{2x^2+1}#
Now we have to compute
#d/dx [4x-3]=d/dx [4x] - d/dx [3]=4d/dx[x]+0=4*1=4#
and the derivative of a square root, which can be written as a power: #sqrt{2x^2+1}=(2x^2+1)^(1/2)#. This is a composition of the square root with a second degree polynomial. The derivative of the composition #a(b(x))# is given by #a'(b(x))*b'(x)#. In this case #a(u)=u^{1/2}# and #b(x)=2x^2+1#, so
#d/{du} [a(u)]=1/2 u^{-1/2}=frac{1}{2sqrt{u}}#
#d/{dx} [b(x)]=d/dx [2x^2] + d/dx [1]=2 d/dx[x^2]+0=2*2x=4x#
So #d/dx [a(b(x))]=frac{1}{2sqrt{2x^2+1}}*4x=frac{2x}{sqrt{2x^2+1}}#.
Finally we get
#d/dx [frac{4x-3}{sqrt{2x^2+1}}]=frac{4*sqrt{2x^2+1}-(4x-3)frac{2x}{sqrt{2x^2+1}}}{2x^2+1}=frac{4(2x^2+1)-(4x-3)2x}{(2x^2+1)^{3/2}}=frac{2(3x+2)}{(2x^2+1)^{3/2}}#
