I just started Calculus and here is a problem I just cannot seem to get the hold of:

"Use the relation 1/(a^{2 }- x^{2}) = 1/2a [1/(a+x) + 1/(a-x)] to find the nth derivative of 1/(a^{2}-x^{2})

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- Nov 23rd 2012, 07:51 AMSupreeth17Calculus: Differentiation
I just started Calculus and here is a problem I just cannot seem to get the hold of:

"Use the relation 1/(a^{2 }- x^{2}) = 1/2a [1/(a+x) + 1/(a-x)] to find the nth derivative of 1/(a^{2}-x^{2}) - Nov 23rd 2012, 10:48 AMMarkFLRe: Calculus: Differentiation
This forum is for introductions, the calculus forum would be the place for this topic. I know you are new here, so I am just offering friendly advice. :)

I would look at the first few derivatives to see if a pattern emerges. We need only work on the factor involving $\displaystyle x$.

$\displaystyle \frac{d}{dx}\left(\frac{1}{a+x}+\frac{1}{a-x} \right)=-\frac{1}{(a+x)^2}+\frac{1}{(a-x)^2}$

$\displaystyle \frac{d}{dx}\left(-\frac{1}{(a+x)^2}+\frac{1}{(a-x)^2} \right)=2\left(\frac{1}{(a+x)^3}+\frac{1}{(a-x)^3} \right)$

$\displaystyle \frac{d}{dx}\left(2\left(\frac{1}{(a+x)^3}+\frac{1 }{(a-x)^3} \right) \right)=6\left(-\frac{1}{(a+x)^4}+\frac{1}{(a-x)^4} \right)$

We may now state the induction hypothesis $\displaystyle P_n$:

$\displaystyle \frac{d^n}{dx^n}\left(\frac{1}{a^2-x^2} \right)=\frac{n!}{2a}\left(\frac{(-1)^n}{(a+x)^{n+1}}+\frac{1}{(a-x)^{n+1}} \right)$

Now, differentiate both sides to try to obtain $\displaystyle P_{n+1}$. - Nov 28th 2012, 02:23 AMSupreeth17Re: Calculus: Differentiation
Thanks ;)