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Find the derivative of f(tanx)wdotrdottg...

Find the derivative of `f(tanx)wdotrdottg(secx)a tx=pi/4,` where `f^(prime)(1)=2a n dg^(prime)(sqrt(2))=4.`

Text Solution

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`"Let "u=f(tan x) and v=g(sec x).` Therefore,
`(du)/(dv)=f'(tan x) sec^(2)x`
`" and "(dv)/(dx)=g'(sec x) sec x tan x`
`therefore" "(du)/(dv)=(du)/(dx)//(dv)/(dx)=(f'(tan x) sec^(2)x)/(g'(sec x) sec x tan x)`
`"or "[(du)/(dv)]_(x=(pi)/(4))=(f'(tan""(pi)/(4)))/(g'(sec""(pi)/(4))sin""(pi)/(4))=(f'(1)sqrt(2))/(g'(sqrt(2)))`
`=(2sqrt(2))/(4)=(1)/(sqrt(2))`
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Statement 1: If both functions f(t)a n dg(t) are continuous on the closed interval [a,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 continuous 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 Lagranges mean value theorem.