Let `f(x) = x^2-4x-3`, `x > 2` and let g be the inverse of f. Find the value of g', where `f(x)= 2`.

Let f(x)=x^(2)-4x-3, x gt2 and g(x) be the inverse of f(x) . Then the value fo (g') , where f(x)=2 , is (here, g' represents the first derivative of g)

Let f(x)=x^(2)-4x-3, x gt2 and g(x) be the inverse of f(x) . Then the value fo (g') , where f(x)=2 , is (here, g' represents the first derivative of g)

Let f(x)=int_2^x (dt)/sqrt(1+t^4) and g be the inverse of f . Then, the value of g'(x) is

Let f(x)=x^(3)+3x+2 and g(x) is inverse function of f(x) then the value of g'(6) is

Let f (x) = x^(3)+ 4x ^(2)+ 6x and g (x) be inverse then the vlaue of g' (-4):

Let f (x) = x^(3)+ 4x ^(2)+ 6x and g (x) be inverse then the vlaue of g' (-4):

Let f(x)=int_(2)^(x)(dt)/(sqrt(1+t^(4))) and g be the inverse of f. Then,the value of g'(x) is

Let f(x)=x^(2)-ax+3a-4 and g(a) be the least value of f(x) then the greatest value of g(a)

Let f(x)=(1)/(1+x^(2)) and g(x) is the inverse of f(x) ,then find g(x)

Let f(x) = x + cos x + 2 and g(x) be the inverse function of f(x), then g'(3) equals _