Prove but translated in English

Tunatumia chain rule.

$$y = \sin(\ln x)$$ $$\frac{dy}{dx} = \cos(\ln x) \cdot \frac{d}{dx}(\ln x)$$

Tunajua kwamba $\frac{d}{dx}(\ln x) = \frac{1}{x}$.

$$\frac{dy}{dx} = \cos(\ln x) \cdot \frac{1}{x}$$ $$\frac{dy}{dx} = \frac{\cos(\ln x)}{x}$$

Tunatumia quotient rule kwa $\frac{dy}{dx} = \frac{\cos(\ln x)}{x}$.
Kumbuka: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.
Hapa, $u = \cos(\ln x)$ na $v = x$.
$u' = \frac{d}{dx}(\cos(\ln x)) = -\sin(\ln x) \cdot \frac{1}{x} = -\frac{\sin(\ln x)}{x}$.
$v' = \frac{d}{dx}(x) = 1$.
Sasa badilisha hizi kwenye fomula ya quotient rule:

$$\frac{d^2y}{dx^2} = \frac{\left(-\frac{\sin(\ln x)}{x}\right) \cdot x - \cos(\ln x) \cdot 1}{x^2}$$ $$\frac{d^2y}{dx^2} = \frac{-\sin(\ln x) - \cos(\ln x)}{x^2}$$

Chukua upande wa kushoto wa equation (LHS):

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

Rahisisha maneno:

$$(-\sin(\ln x) - \cos(\ln x)) + (\cos(\ln x)) + \sin(\ln x)$$

Panga upya maneno:

$$-\sin(\ln x) + \sin(\ln x) - \cos(\ln x) + \cos(\ln x)$$ $$0 + 0 = 0$$

Kwa kuwa upande wa kushoto ni sawa na upande wa kulia (RHS = 0), tumethibitisha equation.