What is the second derivative of f(x)=cos(-1/x^3) ?

Mar 13, 2016

$f ' ' \left(x\right) = \frac{9}{x} ^ 8 \cos \left(- \frac{1}{x} ^ 3\right) - \frac{12}{x} ^ 5 \sin \left(- \frac{1}{x} ^ 3\right)$

Explanation:

For first derivative of $f \left(x\right) = \cos \left(- \frac{1}{x} ^ 3\right)$, we may use function of function formula i.e. $\frac{\mathrm{df} \left(g \left(x\right)\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dx}}$

$\frac{\mathrm{df}}{\mathrm{dx}} = - \sin \left(- \frac{1}{x} ^ 3\right) \times - \left(- 3\right) {x}^{-} 4 = - 3 {x}^{-} 4 \sin \left(- \frac{1}{x} ^ 3\right)$ using $f \left({x}^{-} n\right) = - n {x}^{-} \left(n + 1\right)$

For second derivative, we use the formula of product derivatives i.e.

$\frac{\mathrm{df} \left(x\right) g \left(x\right)}{\mathrm{dx}} = f \left(x\right) g ' \left(x\right) + g \left(x\right) f ' \left(x\right)$, where $f ' \left(x\right)$ and $g ' \left(x\right)$ are first derivatives of $f \left(x\right)$ and $g \left(x\right)$ respectively. Second derivative is mentioned as $f ' ' \left(x\right)$ and $g ' ' \left(x\right)$.

In the given case,
f''(x)=-3d/dx(x^-4sin(-1/x^3))=-3{x^(-4)((-3)/(x^4)cos(-1/x^3)-(-4/(x^5)sin(-1/x^3)} or

$f ' ' \left(x\right) = \frac{9}{x} ^ 8 \cos \left(- \frac{1}{x} ^ 3\right) - \frac{12}{x} ^ 5 \sin \left(- \frac{1}{x} ^ 3\right)$