Given that $y=\arctan(x^2)$ find $\ \dfrac{d^2y}{dx^2}$.
I got
$$\frac{dy}{dx}=\frac{2x}{1+x^4}.$$
Using low d high minus high d low over low squared, I got
$$\frac{d^2y}{dx^2}=\frac{(1+x)^4 \cdot 2 - 2x \cdot 4(1+x)^3}{(1+x^4)^2}.$$
I tried to simplify this but didn't get the answer which is
$$\frac{d^2y}{dx^2}=\frac{2(1-3x^4)}{(1+x^4)^2}.$$
Where am I going wrong?
$\endgroup$ 13 Answers
$\begingroup$You did the low d high minus high d low wrong: The numerator should be:
$$(1+x^4)2 - (2x)(4x^3)$$
$\endgroup$ $\begingroup$Once you get to the second step, you can indeed use the chain rule. Perhaps if you write out what you intend to find, rather than simply finding it, it'll become easier to see:
$$\left(\frac{2x}{1+x^4} \right)' = \frac{(2x)'\left(1+x^4\right) + (2x)\left( 1+x^4\right)'}{\left(1 + x^4 \right)^2}$$
$\endgroup$ $\begingroup$Alternatively,
$ \large \tan (y) = x^2 \Rightarrow \sec^2 (y) \cdot \frac { \mathrm{d}y}{\mathrm{d}x} = 2x $
$ \large \Rightarrow \frac { \mathrm{d}y}{\mathrm{d}x} = \frac {2x}{\sec^2 (y)} = \frac {2x}{\tan^2 (y) + 1} = \frac {2x}{x^4 + 1} $
$ \large \Rightarrow \frac { \mathrm{d^2}y}{\mathrm{d}x^2} = \frac {2(x^4+ 1) - 2x(4x^3)}{(x^4+1)^2} = \frac {-6x^4 + 2}{(x^4 + 1)^2} $
$\endgroup$
