# Calculus I: derivative

• Mar 29th 2009, 03:39 PM
NonCommAlg
Calculus I: derivative
Let $\displaystyle y=\tan^{-1}x.$ Find a closed-form expression for $\displaystyle \frac{d^n y}{dx^n}, \ n \geq 1.$
• Mar 29th 2009, 08:38 PM
dropout_expert
Since $\displaystyle \int^x_0 \frac{dz}{1+z^2} = \tan^{-1}(x)$, we need to find the n-1th derivative of the integrand. Write $\displaystyle \frac{1}{x^2 + 1} = \frac{1}{2i}(\frac{1}{x-i} - \frac{1}{x+i})$ then we differentiate each expression n-1 times to give us $\displaystyle \frac{(-1)^{n-1} (n-1)!}{2i} \left ( \frac{1}{(x-i)^{n}} - \frac{1}{(x+i)^n} \right ).$ Changing to polars and applying De Moivre's gives us: $\displaystyle \frac{1}{(x^2+1)^{\frac{n}{2}}}(n-1)! (-1)^{n-1} \sin(n\tan^{-1}(\frac{1}{x})).$
• Mar 30th 2009, 02:39 AM
NonCommAlg
Quote:

Originally Posted by dropout_expert

Since $\displaystyle \int^x_0 \frac{dz}{1+z^2} = \tan^{-1}(x)$, we need to find the n-1th derivative of the integrand. Write $\displaystyle \frac{1}{x^2 + 1} = \frac{1}{2i}(\frac{1}{x-i} - \frac{1}{x+i})$ then we differentiate each expression n-1 times to give us $\displaystyle \frac{(-1)^{n-1} (n-1)!}{2i} \left ( \frac{1}{(x-i)^{n}} - \frac{1}{(x+i)^n} \right ).$ Changing to polars and applying De Moivre's gives us: $\displaystyle \frac{1}{(x^2+1)^{\frac{n}{2}}}(n-1)! (-1)^{n-1} \sin(n\tan^{-1}(\frac{1}{x})).$

that's a nice approach but you made a mistake in your solution somewhere because your answer is correct only for $\displaystyle x \geq 0.$ another way is to let $\displaystyle x=\tan t, \ y=t, \ \ -\frac{\pi}{2} < t < \frac{\pi}{2}.$ then an easy

induction shows that: $\displaystyle \frac{d^ny}{dx^n}=(n-1)! \sin \left(\frac{n \pi}{2} + nt \right) \cos^n t.$ writing everything in terms of $\displaystyle x$ will give us: $\displaystyle \frac{d^ny}{dx^n}=\frac{(n-1)!}{(x^2 + 1)^{\frac{n}{2}}} \sin \left(\frac{n \pi}{2} + n \tan^{-1}x \right).$