Find $f^{(25)}(x)$ if $f(x)=x^{-3}$ by first finding general solution.

Question

I don't get it what is the question asking for either composition or derivative of f 25th times.

Solution 1.

Sol: $f(x)=x^{-3}=1/x^3$ $f^{(2)}(x)=f(f(x))=f(1/x^3)=1/(1/x^3)^3=1/(1/x^9)=x^9$

$f^{(3)}(x)=f(f^{(2)}(x))=f(x^9)=1/(x^9)^3=1/x^{27}$

$f^{(4)}(x)=f(f^{(3)}(x))=f(1/x^{27})=1/(1/x^{27})^3=1/(1/x^{81})=x^{81}$

$f^{(5)}(x)=f(f^{(4)}(x))=f(x^{81})=1/(x^{81})^3=1/x^{243}$

Solution 2.

Sol:

$f(x)=x^{-3}=1/x^3$

$f^{(2)}(x)=-3/x^4$

$f^{(3)}(x)=12/x^5$

$f^{(4)}(x)=-60/x^6$

$f^{(5)}(x)=360/x^7$

Answer 1

This can be solved via induction.

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Computing the first few derivatives yields a pattern. We deduce $$ I(n):\partial_x^n\frac{1}{x^3}=(-1)^n\frac{(3+n-1)!}{2}\frac{1}{x^{3+n}}. $$ You have already proved the case for $I(1)$. Now assuming $I(n)$ holds for some $n$ we show $$ \partial_x^{n+1}\frac{1}{x^3}=\partial_x(-1)^n\frac{(3+n-1)!}{2}\frac{1}{x^{3+n}}=-(-1)^n\frac{(3+n)(3+n-1)!}{2}\frac{1}{x^{3+n+1}}=(-1)^{n+1}\frac{(3+(n+1)-1)!}{2}\frac{1}{x^{3+n+1}}=I(n+1). $$ Thus, $I(n)\implies I(n+1)$ and we have proven $$ \partial_x^n\frac{1}{x^3}=(-1)^n\frac{(3+n-1)!}{2}\frac{1}{x^{3+n}},\quad n\in\Bbb N. $$ It follows $$ \partial_x^{25}\frac{1}{x^3}=-\frac{(27)!}{2}\frac{1}{x^{28}}. $$

Answer 2

Since $\frac{d}{dx} (a\cdot x^b)= ab\cdot x^{b-1}$ for real $a$ and $b$, we can deduce that $\frac{d^n}{dx^n} (a\cdot x^b) = a(b)(b-1)...(b-(n-1))x^{b-n}$ for a natural number $n$.

So for your problem, $\frac{d^{25}}{dx^{25}} x^{-3} = (-3)(-4)...(-3-24)x^{-3-25}$. You can carry on from there by factoring out $-1$ from each constant factor and writing in factorial form.