Question

How can I find the derivative of the inverse of $f\left( x \right) = {x^3} + x + 1$ at $x = 11$ ?

Answer 1

Hint: We have given a function $f\left( x \right)$ and we have to determine the derivative of the inverse of the function $f\left( x \right)$. To determine the derivative of the inverse of the given function, we use Leibniz rule. According to Leibniz Rule we can write the differentiation of the inverse function as $\dfrac{{dx}}{{dy}} = \dfrac{1}{{\dfrac{{dy}}{{dx}}}}$ . More clearly, in terms of the inverse function it is written as ${\left( {{f^{ - 1}}} \right)^\prime }y = \dfrac{1}{{{f^\prime }\left( {{f^{ - 1}}\left( y \right)} \right)}}$
Differentiation of ${x^n}$ is given as $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$

Complete step by step answer:
Let us consider that the given function is $f\left( x \right) = {x^3} + x + 1$equal to $y$ and also consider that $x = {f^{ - 1}}\left( y \right)$ is the inverse function of $f$. We have to determine the derivative of the given function $f\left( x \right)$ at $x = 11$.
When we substitute $x = 2$ , we get the value of the given function as
$f\left( 2 \right) = {2^3} + 2 + 1$
$ \Rightarrow f\left( 2 \right) = 8 + 2 + 1$
$ \Rightarrow f\left( 2 \right) = 11$
This means that ${f^{ - 1}}\left( {11} \right) = 2$ .

Now we can use the Leibniz rule of differentiation. We need to determine the value of ${\left(f^{ - 1} \right)'}\left( {11} \right)$ which is equal to $\dfrac{1}{{{f^\prime }\left( {{f^{ - 1}}\left( {11} \right)} \right)}}$ .
We know that ${f^{ - 1}}\left( {11} \right) = 2$
$ \Rightarrow {\left(f^{ - 1} \right)'}\left( {11} \right) = \dfrac{1}{{{f^\prime }\left( 2 \right)}}$

Now we determine the derivative of the given function $f\left( x \right) = {x^3} + x + 1$.
$ \Rightarrow {f^\prime }\left( x \right) = 3{x^2} + 1$
Substituting the value of $x = 2$ , we get
$ \Rightarrow {f^\prime }\left( 2 \right) = 3{\left( 2 \right)^2} + 1$
$ \Rightarrow {f^\prime }\left( 2 \right) = 3 \times 4 + 1$
$ \Rightarrow {f^\prime }\left( 2 \right) = 12 + 1$
$ \Rightarrow {f^\prime }\left( 2 \right) = 13$
Step 4: Substituting the value ${f^\prime }\left( 2 \right) = 13$in step (2), we get
$ \Rightarrow {\left( {{f^{ - 1}}} \right)^\prime }\left( {11} \right) = \dfrac{1}{{13}}$

Hence the derivative of the inverse of the given function $f\left( x \right) = {x^3} + x + 1$ at $x = 11$ is $\dfrac{1}{{13}}$.

Note: If $x$ is an independent variable in $f$ then it is a dependent variable in ${f^{ - 1}}$. It is better not to “swap” the variables in writing the inverse function. Notice that we have taken $11$ to be a $y$ value instead of $x$ value, because we did not swap the variables in writing the inverse of the function.