What's the derivative of #f(x) = (arctan (x/2)) + ((5x-1)/(2(x^2) + 4))#?

1 Answer
Sep 3, 2016

#f'(x)=(56+8x+6x^2+2x^3-x^4)/(2(x^2+4)(x^2+2)^2)#.

Explanation:

Let #f(x)=g(x)+h(x)", so that, "f'(x)=g'(x)+h'(x).........(star)#,

where, #g(x)=arc tan (x/2), and, h(x)={(5x-1)/(2x^2+4)}#.

#g(x)=arc tan (x/2)rArrg'(x)=(1/(1+(x/2)^2))*d/dx(x/2)#

#rArr f'(x)=1/2*(4/(4+x^2))=2/(x^2+4).....................(1)#

#h(x)=(5x-1)/(2x^2+4)#

#:.lnh(x)=ln(5x-1)-ln(2x^2+4)#

#:.d/dxlnh(x)=d/dxln(5x-1)-d/dxln(2x^2+4)#

#:. 1/(h(x))d/dxh(x)....................................."[by chain rule]"#

#=1/(5x-1)d/dx(5x-1)-1/(2x^2+4)*d/dx(2x^2+4)#

#=5/(5x-1)-(4x)/(2x^2+4)#

#:. h'(x)=h(x)[{5(2x^2+4)-4x(5x-1)}/{(5x-1)(2x^2+4)}]#

#:.h'(x)=cancel((5x-1))/(2x^2+4){(20+4x-10x^2)/(cancel((5x-1))(2x^2+4))}#

#:. h'(x)=(10+2x-5x^2)/(2(x^2+2)^2).............(2)#

Using #(1),(2)" in "(star)#,

#f'(x)=2/(x^2+4)+(10+2x-5x^2)/(2(x^2+2)^2)#

#={4(x^2+2)^2+(4+x^2)(10+2x-5x^2)}/(2(x^2+4)(x^2+2)^2)#

#=(4x^4+16x^2+16+40+8x-20x^2+10x^2+2x^3-5x^4)/(2(x^2+4)(x^2+2)^2)#

#f'(x)=(56+8x+6x^2+2x^3-x^4)/(2(x^2+4)(x^2+2)^2)#.