Question

if x = log(p), y = 1/p, then d²y/dx² =?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:x = logp$

and

$\rm :\longmapsto\:y = \dfrac{1}{p}$

Now, Consider

$\rm :\longmapsto\:x = logp$

$\rm\implies \:\dfrac{dx}{dp} = \dfrac{1}{p} - - - (1)$

Now,

$\rm :\longmapsto\:y = \dfrac{1}{p}$

$\rm :\longmapsto\:y = {p}^{ - 1}$

$\rm :\longmapsto\:\dfrac{dy}{dp} = - 1 {p}^{ - 1 - 1}$

$\rm :\longmapsto\:\dfrac{dy}{dp} = - {p}^{ - 2}$

$\rm\implies \:\dfrac{dy}{dp} \: = - \: \dfrac{1}{ {p}^{2} }$

So,

$\rm :\longmapsto\:\dfrac{dy}{dx}$

$\rm \:  =  \: \dfrac{dy}{dp} \div \dfrac{dx}{dp}$

$\rm \:  =  \: - \dfrac{1}{ {p}^{2} } \div \dfrac{1}{p}$

$\rm \:  =  \: - \dfrac{1}{ p }$

$\bf\implies \:\dfrac{dy}{dx} \: = \: - \: \dfrac{1}{p}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = - \dfrac{d}{dx} {p}^{ - 1}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = {p}^{ - 1 - 1} \dfrac{dp}{dx}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = {p}^{ - 2} \times p \: \: \: \: \: \: \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = {p}^{ - 2 + 1}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = {p}^{ - 1}$

$\bf\implies \:\dfrac{ {d}^{2}y }{d {x}^{2} } = \dfrac{1}{p}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:x = logp$

and

$\rm :\longmapsto\:y = \dfrac{1}{p}$

Now, Consider

$\rm :\longmapsto\:x = logp$

$\rm\implies \:\dfrac{dx}{dp} = \dfrac{1}{p} - - - (1)$

Now,

$\rm :\longmapsto\:y = \dfrac{1}{p}$

$\rm :\longmapsto\:y = {p}^{ - 1}$

$\rm :\longmapsto\:\dfrac{dy}{dp} = - 1 {p}^{ - 1 - 1}$

$\rm :\longmapsto\:\dfrac{dy}{dp} = - {p}^{ - 2}$

$\rm\implies \:\dfrac{dy}{dp} \: = - \: \dfrac{1}{ {p}^{2} }$

So,

$\rm :\longmapsto\:\dfrac{dy}{dx}$

$\rm \:  =  \: \dfrac{dy}{dp} \div \dfrac{dx}{dp}$

$\rm \:  =  \: - \dfrac{1}{ {p}^{2} } \div \dfrac{1}{p}$

$\rm \:  =  \: - \dfrac{1}{ p }$

$\bf\implies \:\dfrac{dy}{dx} \: = \: - \: \dfrac{1}{p}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = - \dfrac{d}{dx} {p}^{ - 1}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = {p}^{ - 1 - 1} \dfrac{dp}{dx}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = {p}^{ - 2} \times p \: \: \: \: \: \: \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = {p}^{ - 2 + 1}$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = {p}^{ - 1}$

$\bf\implies \:\dfrac{ {d}^{2}y }{d {x}^{2} } = \dfrac{1}{p}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$