Q.If $y=e^x\sin x$, find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$, and show that $\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$.

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 $x$:

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

Now $\dfrac{d}{dy}\!\left(\dfrac{dx}{dy}\right)=\dfrac{d^2x}{dy^2}=m$ and $\dfrac{dy}{dx}=\dfrac1l$, so

$\frac{d^2y}{dx^2} = -\frac{1}{l^2}\cdot m\cdot\frac1l = -\frac{m}{l^3}.$

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 $dy/dx = y/x$:

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

This makes sense geometrically: $dy/dx = y/x$ describes the family of straight lines $y=kx$ through the origin, and a straight line always has zero second derivative (zero curvature).

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 $3(3x+1)^{-1}$):

$f''(x) = 3\cdot(-1)(3x+1)^{-2}\cdot 3 = \dfrac{-9}{(3x+1)^2}$

Substitute $x=1$:

$f''(1) = \dfrac{-9}{(3(1)+1)^2} = \dfrac{-9}{4^2} = \dfrac{-9}{16}$