Answer to Question #103971 in Calculus for Martin

Question

Find d²y/dx² for y=csc(x)

Answer 1

Solution:

1.      This expression can be written as:

$\frac{d^2y}{d^2x}(\csc(x)) = \frac{dy}{dx}(\frac{dy}{dx}\csc(x))$

2.      Find the first derivative of the specified function:

$\frac{dy}{dx}\csc(x) = \frac{dy}{dx}(\frac{1}{\sin(x)}) = \frac{(1^{\prime})×\sin(x) - 1×(\sin(x))^{\prime}}{\sin^{2}(x)} = - \frac{\cos(x)}{\sin^{2}(x)}$

3.      Next, find the derivative of the resulting function:

$\frac{dy}{dx}(\frac{-\cos(x)}{\sin^{2}(x)}) = \frac{(-\cos(x))^{\prime}×\sin^{2}(x) - (-\cos(x))×(\sin^{2}(x))^{\prime}}{(\sin^{2}(x))^{2}} =$

$= \frac{\sin(x) × \sin^{2}(x) + \cos(x) × (2 \sin(x)\cos(x))}{\sin^{4}(x)} = \frac{\sin(x)×(\sin^{2}(x) + 2\cos^{2}(x))}{\sin^{4}(x)} =$

$= \frac{\sin^{2}(x) + \cos^{2}(x) +\cos^{2}(x) }{\sin^{3}(x)} = \frac{1+ \cos^{2}(x)}{\sin^{3}(x)} = \frac{1}{\sin^{3}(x)}(1 + \cos^{2}(x)) =$

$= \csc^{3}(x)(1+\cos^{2}(x))$

Answer:

 $\frac{d^2y}{d^2x}(\csc(x)) = (\cos^{2}(x) + 1) \csc^{3}(x)$