Let `g(x)` be the inverse of an invertible function `f(x),` which is differentiable for all real `xdot` Then `g^('')(f(x))` equals. (a)`-(f^('')(x))/((f^'(x))^3)` (b) `(f^(prime)(x)f^('')(x)-(f^(prime)(x))^3)/(f^(prime)(x))` (c)`(f^(prime)(x)f^('')(x)-(f^(prime)(x))^2)/((f^(prime)(x))^2)` (d) none of these

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

Solve x dy=(y+x(f(y/x))/(f^(prime)(y/x)))dx

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))

Suppose fa n dg are functions having second derivative f'' and g' ' everywhere. If f(x)dotg(x)=1 for all xa n df^(prime)a n dg' are never zero, then (f^('')(x))/(f^(prime)(x))-(g^('')(x))/(g^(prime)(x))e q u a l (a) (-2f^(prime)(x))/f (b) (2g^(prime)(x))/(g(x)) (c) (-f^(prime)(x))/(f(x)) (d) (2f^(prime)(x))/(f(x))

Evaluate: inte^x(f(x)+f^(prime)(x))dx=e^xf(x)+C

Using the first principle, prove that d/(dx)(1/(f(x)))=(-f^(prime)(x))/([f(x)]^2)

If f(x-y),f(x)f(y),a n df(x+y) are in A.P. for all x , y ,a n df(0)!=0, then (a) f(4)=f(-4) (b) f(2)+f(-2)=0 (c) f^(prime)(4)+f^(prime)(-4)=0 (d) f^(prime)(2)=f^(prime)(-2)

Let y=f(x) be a parabola, having its axis parallel to the y-axis, which is touched by the line y=x at x=1. Then, (a) 2f(0)=1-f^(prime)(0) (b) f(0)+f^(prime)(0)+f^(0)=1 (c) f^(prime)(1)=1 (d) f^(prime)(0)=f^(prime)(1)

A function f: Rvec[1,oo) satisfies the equation f(x y)=f(x)f(y)-f(x)-f(y)+2. If differentiable on R-{0}a n df(2)=5,f^(prime)(x)=(f(x)-1)/xdotlambdat h e nlambda= 2^(prime)f(1) b. 3f^(prime)(1) c. 1/2f^(prime)(1) d. f^(prime)(1)

Let f: R->R satisfying |f(x)|lt=x^2,AAx in R be differentiable at x=0. Then find f^(prime)(0)dot