Thread: Differentiation

1. Differentiation

For the curve with equation y = arsinh (x + 1), find
(a) the coordinates of its point of inflexion P
(b) an equation of the normal to the curve at P.

For (a), I know that if dy/dx = 0 and d^2 y/dx^2 = 0, but d^3 y/dx^3 ≠ 0, then the point is a point of inflexion. But now how can I start to solve this?

For (b), after finding P, it may be easy.

2. Originally Posted by geton
For the curve with equation y = arsinh (x + 1), find
(a) the coordinates of its point of inflexion P
(b) an equation of the normal to the curve at P.

For (a), I know that if dy/dx = 0 and d^2 y/dx^2 = 0, but d^3 y/dx^3 ≠ 0, then the point is a point of inflexion. But now how can I start to solve this?

For (b), after finding P, it may be easy.
You start by getting dy/dx ..... Can you do that?

3. Originally Posted by mr fantastic
You start by getting dy/dx ..... Can you do that?

Yes. Its done. Thanks.

4. Originally Posted by geton
Yes. Its done. Thanks.
Right then.

$\frac{dy}{dx} = \frac{1}{\sqrt{1 + (x + 1)^2}} = \frac{1}{\sqrt{x^2 + 2x + 2}} = (x^2 + 2x + 2)^{-1/2}$

$\Rightarrow \frac{d^2y}{dx^2} = -\frac{1}{2} (x^2 + 2x + 2)^{-3/2} \, (2x + 2) = -\frac{1}{2} \, \frac{2x+2}{(x^2 + 2x + 2)^{3/2}}$.

$\frac{d^2y}{dx^2} = 0 \Rightarrow 2x+2 = 0 \Rightarrow x = -1$.

Rather than calculating the third derivative, you can test the nature of this solution by examining whether it corresponds to a turning point of $\frac{dy}{dx}$. This is easily done using the usual sign test .....

5. Originally Posted by geton
For the curve with equation y = arsinh (x + 1), find
(a) the coordinates of its point of inflexion P
[snip]
Alternatively, you could use the 'well known' fact that y = arsinh(x) has a point of inflexion at (0, 0) and then apply a translation of 1 unit to the left .....

6. Originally Posted by mr fantastic
Alternatively, you could use the 'well known' fact that y = arsinh(x) has a point of inflexion at (0, 0) and then apply a translation of 1 unit to the left .....
Ya its good too.