Let `f(x)a n dg(x)` be differentiable function in `(a , b),` continuous at `aa n db ,a n dg(x)!=0` in `[a , b]dot` Then prove that `(g(a)f(b)-f(a)g(b))/(g(c)f^(prime)(c)-f(c)g^(prime)(c))=((b-a)g(a)g(b))/((g(c))^2)`

Let f(x) and g(x) be differentiable function in (a,b), continuous at a and b, and g(x)!=0 in [a,b]. Then prove that (g(a)f(b)-f(a)g(b))/(g(c)f'(c)-f(c)g'(c))=((b-a)g(a)g(b))/((g(c))^(2))

Statement 1: If both functions f(t)a n dg(t) are continuous on the closed interval [1,b], differentiable on the open interval (a,b) and g^(prime)(t) is not zero on that open interval, then there exists some c in (a , b) such that (f^(prime)(c))/(g^(prime)(c))=(f(b)-f(a))/(g(b)-g(a)) Statement 2: If f(t)a n dg(t) are continuou and differentiable in [a, b], then there exists some c in (a,b) such that f^(prime)(c)=(f(b)-f(a))/(b-a)a n dg^(prime)(c)(g(b)-g(a))/(b-a) from Lagranes mean value theorem.

f and g differentiable functions of x such that f(g(x))=x," If "g'(a)ne0" and "g(a)=b," then "f'(b)=

If f and g are derivable function of x such that g'(a)ne0,g(a)=b" and "f(g(x))=x," then 'f'(b)=

If the function f(x) and g(x) are continuous in [a, b] and differentiable in (a, b), then the f(a) f (b) equation |(f(a),f(b)),(g(a),g(b))|=(b-a)|(f(a),f'(x)),(g(a),g'(x))| has, in the interval [a,b] :