# Proving $y=\tan(x)$ is of the form $P_{n+1}(\tan(x))$

So as the title states I have to prove using induction that the nth derivative of $y=\tan(x)$ is of the form $P_{n+1}(\tan(x))$, where $P_{n+1}$ is a polynomial of degree $n+1$

So what's the intuition behind this? Usually I would like to find a general formula for the derivative of $\tan(x)$ and then following the steps of mathematical induction. Which would get me:

$n = k+1$

for all n.

How do I approach this problem?

Let $f(x)=\tan x$; the base step is $f'(x)=1+\tan^2x$, which is a polynomial in $\tan x$ of degree $2$.

Suppose $$f^{(n)}(x)=P(\tan x)$$ where $P(X)$ is a polynomial of degree $n+1$. Then, by the chain rule, $$f^{(n+1)}(x)=(1+\tan^2x)P'(\tan x)$$ Note that the degree of $P'(X)$ is $n$, so if $Q(X)=(1+X^2)P'(X)$, then the degree of $Q(X)$ is $n+2$ and $f^{(n+1)}(x)=Q(\tan x)$.

• It was very helpful. Although, regarding the degrees. Do you consider "n" as the first degree, i.e the first derivative will be the second degree? Becasue should't $P'(tan x)$ have the degree $n+1$? Nov 26, 2017 at 23:07
• @AliasaZarownyPseudonymia $P'$ means the derivative of $P$. In my opinion adding notation, such as an index showing the degree, is of a hindrance in understanding what happens. Nov 26, 2017 at 23:08

We have $y= \tan x$ so \begin{eqnarray*} \frac{dy}{dx} =\sec^2 x =1+\tan^2 x =p_2(\tan x). \end{eqnarray*} Now assume \begin{eqnarray*} \frac{d^{n}y}{dx^{n} } =\sum_{r=0}^{n+1} a_{n,r} \tan^r x =p_{n+1}(\tan x) \end{eqnarray*} (where $a_{n,r}$ are constants) i.e a polynomial in $\tan x$. Differentiating gives \begin{eqnarray*} \frac{d^{n+1}y}{dx^{n+1} } =\sum_{r=1}^{n+1} r a_{n,r} \tan^{r-1} x (1+\tan^2 x) =p_{n+2}(\tan x) \end{eqnarray*} which is clearly a polynomial in $\tan x$.

• The key step here is to notice that the derivative $\sec^2 x$ of $\tan x$ is itself a polynomial in $\tan x$. The rest of the proof follows easily once you notice this. Nov 26, 2017 at 22:11
• the assumption, is that an assumption for a general formula for the derivative of tan(x)? Nov 26, 2017 at 22:29
• Yes, we are assuming that the $n^{th}$ derivative is a polynomial of degree $n+1$. Nov 26, 2017 at 22:33
• Ah. However i'm not really seeing how the last answer equates to $p_{n+1}tan(x)$ ? I can see that $p_{n+1}$ is the function of the first derivative which in this case is $1+tan^2 x$ but is $ra_{n,r}tan^r-1 x$ to be equal to tan(x)? Nov 26, 2017 at 22:49