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# If the function $f(x) = a + bx$ and ${f^r} = fff...$(repeated $r$ times), then $\frac{d}{{dx}}\left\{ {{f^r}(x)} \right\}$ is equal toE.$a + {b^r}x$F.$ar + {b^r}x$G.$ar$H.${b^r}$

Last updated date: 14th Mar 2023
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Hint: Here we proceed by observing the logic given in the question.
Given, $f(x) = a + bx$
$f\left\{ {f(x)} \right\} = a + b(a + bx)$
$= ab + a + {b^2}x$
$= a(1 + b) + {b^2}x$
$f\left[ {f\left\{ {f(x)} \right\}} \right] = f\left\{ {a(1 + b) + {b^2}x} \right\}$
$= a + b\left\{ {a(1 + b) + {b^2}x} \right\}$
$= a(1 + b + {b^2}) + {b^3}x$
Therefore, ${f^r}(x) = a(1 + b + {b^2} + .... + {b^{r - 1}}) + {b^r}x$
$${f^r}x = a\left( {\frac{{{b^r} - 1}}{{b - 1}}} \right] + {b^r}x$$
$\Rightarrow \frac{d}{{dx}}\left\{ {{f^r}x} \right\} = {b^r}$

Note: The given function is$f(x) = a + bx$, where x is variable. We are finding a general form of ${f^r}(x)$from$f(x)$. Then differentiating the obtained ${f^r}(x)$ with respect to x to get the required answer.