How do you differentiate f(x)= cosx/ (sinx) twice using the quotient rule?

Dec 4, 2015

$f ' \left(x\right) = - {\csc}^{2} x , f ' ' \left(x\right) = 2 {\csc}^{2} x \cot x$

Explanation:

Through the quotient rule:

$f ' \left(x\right) = \frac{\sin x \frac{d}{\mathrm{dx}} \left[\cos x\right] - \cos x \frac{d}{\mathrm{dx}} \left[\sin x\right]}{\sin} ^ 2 x$

Recall that:

$\frac{d}{\mathrm{dx}} \left[\cos x\right] = - \sin x$

$\frac{d}{\mathrm{dx}} \left[\sin x\right] = \cos x$

$f ' \left(x\right) = \frac{- {\sin}^{2} x - {\cos}^{2} x}{\sin} ^ 2 x$

$f ' \left(x\right) = - \frac{{\sin}^{2} x + {\cos}^{2} x}{\sin} ^ 2 x$

$f ' \left(x\right) = - \frac{1}{\sin} ^ 2 x$

$f ' \left(x\right) = - {\csc}^{2} x$

In order to find the second derivative, if we want to continue using the quotient rule, use:

$f ' \left(x\right) = \frac{- 1}{\sin} ^ 2 x$

$f ' ' \left(x\right) = \frac{{\sin}^{2} x \frac{d}{\mathrm{dx}} \left[- 1\right] - \left(- 1\right) \frac{d}{\mathrm{dx}} \left[{\sin}^{2} x\right]}{\sin} ^ 4 x$

Again, find each derivative:

$\frac{d}{\mathrm{dx}} \left[- 1\right] = 0$

Chain rule coming up:

$\frac{d}{\mathrm{dx}} \left[{\sin}^{2} x\right] = 2 \sin x \frac{d}{\mathrm{dx}} \left[\sin x\right] = 2 \sin x \cos x$

Plug back in:

$f ' ' \left(x\right) = \frac{2 \sin x \cos x}{\sin} ^ 4 x$

$f ' ' \left(x\right) = \frac{2 \cos x}{\sin} ^ 3 x$

$f ' ' \left(x\right) = 2 \left(\frac{1}{\sin} ^ 2 x\right) \left(\cos \frac{x}{\sin} x\right)$

$f ' ' \left(x\right) = 2 {\csc}^{2} x \cot x$