Let f(x) = (2x-pie) 3 +2x - cosx. Then show that d[ f -1 (x)] / dx at x=pie/2 is 1/3.

Question

Ramesh V · Answer

(1) It's "Pi". and Pi is not 22/7

first find df(x) / dx.

df(x)/dx = 3(2x - pi)² * 2 + 2 + sin(x)

=6(2x - pi)² + 2 + sin(x)

Let x = pi/2 , we get: 0 + 2 + 1 = 3

So the answer for df^-1(x)/dx i.e., 1/[df(x)/dx] is : 1/3

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Let f(x) = (2x-pie) 3 +2x - cosx. Then show that d[ f -1 (x)] / dx at x=pie/2 is 1/3.

Let f(x) = (2x-pie)³+2x - cosx. Then show that d[ f ^-1 (x)] / dx at x=pie/2 is 1/3.

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