How do you differentiate f(x)= cscx twice using the quotient rule?

1 Answer
Mar 16, 2018

f''(x) = 2csc^(3)(x) - csc(x)

Explanation:

We have: f(x) = csc(x) = frac(1)(sin(x))

Rightarrow f'(x) = frac(frac(d)(dx)(1) cdot sin(x) - frac(d)(dx)(sin(x)) cdot 1)(sin^(2)(x))

Rightarrow f'(x) = frac(0 cdot sin(x) - cos(x) cdot 1)(sin^(2)(x))

Rightarrow f'(x) = - frac(cos(x))(sin^(2)(x))

Rightarrow f'(x) = - frac(cos(x))(sin(x)) cdot frac(1)(sin(x))

Rightarrow f'(x) = - cot(x) cdot frac(1)(sin(x))

Rightarrow f'(x) = - frac(cot(x))(sin(x))

Then, let's differentiate f'(x):

Rightarrow f''(x) = - (frac(frac(d)(dx)(cot(x)) cdot sin(x) - frac(d)(dx)(sin(x)) cdot cot(x))(sin^(2)(x)))

Rightarrow f''(x) = - (frac(- csc^(2)(x) cdot sin(x) - cos(x) cdot cot(x))(sin^(2)(x)))

Rightarrow f''(x) = - (frac(- (csc^(2)(x) sin(x) + cos(x) cot(x)))(sin^(2)(x)))

Rightarrow f''(x) = frac(csc^(2)(x) sin(x) + cos(x) cot(x))(sin^(2)(x))

Rightarrow f''(x) = frac(csc^(2)(x) sin(x))(sin^(2)(x)) + frac(cos(x) cot(x))(sin^(2)(x))

Rightarrow f''(x) = frac(csc^(2)(x))(sin(x)) + frac(cos(x))(sin(x)) cdot frac(cot(x))(sin(x))

Rightarrow f''(x) = csc^(2)(x) cdot csc(x) + cot(x) cdot cot(x) cdot csc(x)

Rightarrow f''(x) = csc^(3)(x) + cot^(2)(x) csc(x)

One of the Pythagorean identities is cos^(2)(x) + sin^(2)(x) = 1.

If we divide through by sin^(2)(x), we get:

Rightarrow frac(cos^(2)(x))(sin^(2)(x)) + frac(sin^(2)(x))(sin^(2)(x)) = frac(1)(sin^(2)(x))

Rightarrow 1 + cot^(2)(x) = csc^(2)(x)

Rightarrow cot^(2)(x) = csc^(2)(x) - 1

Let's apply this rearranged identity:

Rightarrow f''(x) = csc^(3)(x) + (csc^(2)(x) - 1) csc(x)

Rightarrow f''(x) = csc^(3)(x) + csc^(3)(x) - csc(x)

therefore f''(x) = 2csc^(3)(x) - csc(x)