IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

sarvpriye

f(x)= (2x-

)³ - cosx +2x,

g(x)= f^-1(x)

Find d/dx (g(x))

siyswarya

Ithink it is 48x-12pi+cosx

elessar_iitkgp

y = (2x -

)³+2x-cosx
dy/dx = 6(2x -

)²+2+sinx

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

Now, f(g) = (2g -

)³+2g-cosg
f'(g) = 6(2g -

)²+2+sing

Hence, dg/dx = 1/[6(2g -

)²+2+sing]

If in this problem you need the value of this derivative at x =

, then see this post
http://www.goiit.com/posts/list/0/differenciation-differention-again-31586.htm#192558

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