If $f(x) =x^(-1/3)$ , what is the derivative of the inverse of f(x)?

Question

the answer is $-3x^-4$ , but i don't know how to get there.

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Answer 1 · 2018-08-13T08:39:01

$d/(dx)(f^-1(x))=-3x^-4$

Here ,

$f(x)=y=x^(-1/3)$

**Let , the function f be one-one and onto.

So , the inverse function of $f(x)$ exists.**

$:.y=x^(-1/3)=(1/x)^(1/3)to[because a^-n=1/a^n]$

$:.y^3=((1/x)^(1/3))^3$

$:.y^3=(1/x)^(1/3xx3) to[because(a^m)^n=a^(mn)]$

$:.y^3=1/x$

$:.x=1/y^3$

$:.color(red)(x=y^(-3)$

We know that ,

$f(x)=y=>x=f^-1(y)to[because "definition of inverse function"]$

$:.$ :.d/(dx)(f^-1(x))=-3x^-4 $x=f^-1(y)=y^-3$

Now changing variable from $yto x,$

$:.color(blue)(f^-1(x)=x^-3$

$:.d/(dx)(f^-1(x))=d/(dx)(x^-3)$

$:.d/(dx)(f^-1(x))=-3x^(-3-1)$

$:.d/(dx)(f^-1(x))=-3x^-4$