`f(x)={(x-1)/(2x^2-7x+5), x!=1 -1/3, x=1,t h e nf^(prime)(1)`

If f(x)={(x-1)/(2x^(2)-7x+5), when x!=1-(1)/(3), when x=1, prove that f'(1)=-(2)/(9)

Function f(x) is defined as f(x)={(x-1)/(2x^(2)-7x+5),x!=1 and -(1)/(3),x=1 .Is f(x) differentiable at x=1 if yes find f'(1)?

Find the derivative of f(x)={(x-1)/(2x^2-7x+5)w h e nx!=1 and-1/3w h e nx=1} atx=1.

Find the derivative of f(x)={(x-1)/(2x^2-7x+5)w h e nx!=1=1/3w h e nx=1a tx=1.

If f(x)=(x-1)/(2x^(2)-7x+5) for x!=1 and f(x)=-(1)/(3) for x=1 then f'(1)=

If f(x)=sqrt(x^2+1),\ \ g(x)=(x+1)/(x^2+1) and h(x)=2x-3 , then find f^(prime)(h^(prime)(g^(prime)(x))) .

If f (x) = (x -1)/( 2x ^(2) - 7x + 5) for x ne 1, f (1) = (-1)/(3), find f '(1).

If f(x) ={{:((x-1)/(2x ^(2) -7x +5),"when"x ne1),(-1/3, "when" x=1):} then the value of f '(1) is-