IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

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y = (2x -

)³+2x-cosx
dy/dx = 6(2x -

)²+2+sinx

Let z = g(x) = f^--1(x)
We need to find dz/dx
x = f(z)
Differentiating wrt x,
1 = f'(z) (dz/dx)
where f'(z) represents the derivative of f(z) wrt z
dz/dx = 1/f'(z)

Now, f(z) = (2z -

)³+2z-cosz
f'(z) = 6(2z -

)²+2+sinz

Hence, dz/dx = 1/[6(2z -

)²+2+sinz]
Now, dz/dx| _{x = pi} = |1/[6(2z -

)²+2+sinz]|_x=pi
So we must find the value of z at at which f(z) is

Now, by inspection we can see that the required value of z is

/2
Hence,
dz/dx| _{x = pi} = |1/[6(2z -

)²+2+sinz]|_z=pi/2 = 1/3

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