Higher-order derivatives

Solution :

Since

$$g(x)=f^{-1}(x)$$

we use inverse function differentiation formulas.

First derivative : $$g'(y)=\frac1{f'(x)}$$ where $$y=f(x)$$

At $$x=1$$ we have $$y=f(1)=2$$

Therefore,

$$g'(2)=\frac1{f'(1)}$$

$$=\frac12$$

Now second derivative :

$$g''(y)=-\frac{f''(x)}{(f'(x))^3}$$

Hence,

$$g''(2) = -\frac{f''(1)}{(f'(1))^3}$$

$$= -\frac3{2^3}$$

$$= -\frac38$$

Now for third derivative :

$$g'''(y) = \frac{3(f''(x))^2-f'(x)f'''(x)} {(f'(x))^5}$$

Substituting given values :

$$g'''(2) = \frac{3(3)^2-(2)(4)}{(2)^5}$$

$$= \frac{27-8}{32}$$

$$= \frac{19}{32}$$

Final Answer :

$$\frac{19}{32}$$

Multiply numerator and denominator by $$\left(\sqrt{1+\sin x} - \sqrt{1-\sin x}\right)$$:

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

$$\left(\sqrt{1+\sin x} - \sqrt{1-\sin x}\right)^2 = 2 - 2\sqrt{1-\sin^2 x} = 2 - 2\cos x$$

Denominator $$= 2\sin x$$. So:

$$f(x) = \frac{2-2\cos x}{2\sin x} = \frac{1-\cos x}{\sin x}$$

Apply half-angle identities $$1-\cos x = 2\sin^2\dfrac{x}{2}$$ and $$\sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}$$:

$$f(x) = \frac{2\sin^2\dfrac{x}{2}}{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}} = \tan\frac{x}{2}$$

$$g(x) = \ln\left(\tan\frac{x}{2}\right)$$

$$g'(x) = \frac{1}{\tan\dfrac{x}{2}} \cdot \sec^2\frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}} = \frac{1}{\sin x} = \cosec x$$

$$g''(x) = \frac{d}{dx}(\cosec x) = \boxed{-\cosec x\cot x}$$