find dy/dx of the following function?

Question

y= (x+1)/(x^2 -1) + (2x+3)((1/x^2)+(1/x))

Answer 1

(1/x-1)+(2x+3)(x+1)/x^2
[(x-1)^(-1)]+(2x^2+5x+3)/x^2
dy/dx=-(x-1)+[(4x+5)*x^2-2x*(2x^2+5x+3)]/x^4
=(5x^2+x-6)/x

Answer 2

♠ thus y= (x+1)/(x^2 -1) + (2x+3)((1/x^2)+(1/x)) =
=1/(x-1) + (2x+3)(1/x^2 + 1/x) = u(x)+v(x); y’=u’+v’;
♣ u’= -1/(x-1)^2;
v’= (2x+3)’*(1/x^2 + 1/x) + (2x+3)(1/x^2 + 1/x)’=
= 2/x^2 +2/x +(2x+3)(-2/x^3) +(2x+3)(-1/x^2) =
= (2x+2x^2 –4x –6 –2x^2 –3x)/x^3 = -(5x+6)/x^3;
♦ y’=-1/(x-1)^2 -(5x+6)/x^3; or
y’= {-x^3 - (5x+6)(x^2-2x+1)}/{x^3 (x-1)^2}=
= {-6x^3 +4x^2 +7x –6}/{x^3 (x-1)^2};

Answer 3

I won't do this for you -- but will tell you how to do it yourself.

The right hand side is the sum of three expressions. The derivative of a sum is equal to the sum of the derivatives. So, you need to find the derivatives of each piece & add them together.

The first equation is the quotient of two polynomials. You need to use the quotient rule to find this derivative. You also need to use the fact that if y=ax^n then dy/dx = n*ax^(n-1)

If you multiply out the second piece, it also becomes the quotient of two polynomials -- so use the quotient rule again.

The last piece is easy. 1/x = x^(-1) -- so its derivative is -x^(-2)

Answer 4

y = (x+1)/(x^2-1) +(2x+3)/x^2 +1/x

dy = ((x+1)2x -(x^2-1))/(x^2-1)^2
+ (2x(2x+3)-2x^2)/x^2
-1/x^2
dy =( x^2+2x+1)/(x^2-1)^2
+(2x^2 +6x-1)/x^2


if you work at it you can probably simplify this