Q.If $x^2+y^2=r^2$ ($r$ constant), find $\dfrac{d^2y}{dx^2}$.

Use the reciprocal relation between $\tfrac{dy}{dx}$ and $\tfrac{dx}{dy}$, then apply the standard second-derivative inversion formula.

Relating $\tfrac{d^2y}{dx^2}$ to $\tfrac{d^2x}{dy^2}$ is a CBSE/NCERT Class 12 continuity and differentiability result.

Since $\dfrac{dx}{dy}=l$, we have $\dfrac{dy}{dx}=\dfrac{1}{l}$.

Differentiate $\dfrac{dy}{dx}=\left(\dfrac{dx}{dy}\right)^{-1}$ with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dx}{dy}\right)^{-1} = -\left(\frac{dx}{dy}\right)^{-2}\frac{d}{dy}\!\left(\frac{dx}{dy}\right)\cdot\frac{dy}{dx}.$ …

Differentiate $dy/dx = y/x$ implicitly using the quotient rule to get $d^2y/dx^2$.

Given $\dfrac{dy}{dx} = \dfrac{y}{x}$. Differentiate both sides with respect to $x$, using the quotient rule on the right side (remembering $y$ is a function of $x$):

$\frac{d^2y}{dx^2} = \frac{x\cdot \dfrac{dy}{dx} - y\cdot 1}{x^2}$

Substitute $dy/dx = y/x$:

$\frac{d^2y}{dx^2} = \frac{x\cdot(y/x) - y}{x^2} = \frac{y-y}{x^2} = 0$

…

Differentiate $f(x)=\log(3x+1)$ twice using the chain rule, then substitute $x=1$.

Given $f(x) = \log(3x+1)$.

First derivative (chain rule, $\dfrac{d}{dx}\log u = \dfrac{1}{u}\cdot\dfrac{du}{dx}$ with $u=3x+1$):

$f'(x) = \dfrac{3}{3x+1}$

Second derivative (quotient/chain rule on $3(3x+1)^{-1}$): …