If `f(x)=x+tanxa n df` is the inverse of `g,` then `g^(prime)(x)` equals (a)`1/(1+[g(x)-x]^2)` (b) `1/(2-[g(x)-x]^2)` (c)`1/(2+[g(x)-x]^2)` (d) none of these

If g is the inverse function of fa n df^(prime)(x)=sinx ,t h e ng^(prime)(x) is (a) cos e c{g(x)} (b) "sin"{g(x)} (c) -1/("sin"{g(x)}) (d) none of these

If f(x)=2x+1 and g(x)=x^2-2 , then gof(x) is

Let g(x) be the inverse of an invertible function f(x) which is differentiable at x=c . Then g^(prime)(f(x)) equal. (a) f^(prime)(c) (b) 1/(f^(prime)(c)) (c) f(c) (d) none of these

If f(x)=x^(3)+3x+4 and g is the inverse function of f(x), then the value of (d)/(dx)((g(x))/(g(g(x)))) at x = 4 equals

If f(x) =x^(2) -2,g (x) =2x+ 1 then f^(@) g (x)

Ifintxlog(1+1/x)dx=f(x)log(x+1)+g(x)x^2+A x+C , then (a) f(x)=1/2x^2 (b) g(x)=logx (c) A=1 (d) none of these

If g is the inverse of a function f and f^'(x)=1/(1+x^5) then g(x) is equal to (1) 1""+x^5 (2) 5x^4 (3) 1/(1+{g(x)}^5) (4) 1+{g(x)}^5

If G(x)=-sqrt(25-x^2),t h e n lim_(x->1)(G(x)-G(1))/(x-1)i s (a) 1/(24) (b) 1/5 (c) -sqrt(24) (d) none of these

Let g^(prime)(x)>0a n df^(prime)(x) g(f(x-1)) f(g(x+1))>f(g(x-1)) g(f(x+1))

If f(x)=1/x ,g(x)=1/(x^2), and h(x)=x^2 , then (A) f(g(x))=x^2,x!=0,h(g(x))=1/(x^2) (B) h(g(x))=1/(x^2),x!=0,fog(x)=x^2 (C) fog(x)=x^2,x!=0,h(g(x))=(g(x))^2,x!=0 (D) none of these