Question

Q1.y=2x+3÷4x+5 find y"​(find the double derivative of y)​.

Answer 1

Step-by-step explanation:

Answer:

$\frac{16}{(4x+5)^3}$

Step-by-step explanation:

Given,

$\begin{gathered}y=\frac{2x+3}{4x+5}\\\;\\y=\frac{2(2x+3)}{2(4x+5)}\\\;\\y=\frac{4x+6}{2(4x+5)}\\\;\\y=\frac{4x+5+1}{2(4x+5)}\\\;\\y=\frac{4x+5}{2(4x+5)}+\frac{1}{2(4x+5)}\\\;\\y=\frac{1}{2}+\frac{1}{2(4x+5)}\\\;\\y=\frac{1}{2}+\frac{(4x+5)^{-1}}{2}\\\;\\\text{Diff. both sides with respect to x}\\\;\\\frac{dy}{dx}=0-\frac{(4x+5)^{-1-1}}{2}.\frac{d(4x+5)}{dx}\\\;\\\frac{dy}{dx}=-\frac{(4x+5)^{-2}}{2}.4\\\;\\\frac{dy}{dx}=-2(4x+5)^{-2}\\\;\\\text{Diff. both sides with respect to x}\end{gathered}$

$\begin{gathered}\\\;\\\frac{d^2y}{dx^2}=-2.(-2).(4x+5)^{-2-1}.\frac{d(4x+5)}{dx}\\\;\\\frac{d^2y}{dx^2}=4(4x+5)^{-3}.4\\\;\\\frac{d^2y}{dx^2}=\frac{16}{(4x+5)^3}\end{gathered}$

Answer 2

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

Given function is

$\rm :\longmapsto\:y = \dfrac{2x + 3}{4x + 5}$

can be rewritten as

$\rm :\longmapsto\:y = \dfrac{1}{2} \bigg[\dfrac{4x + 6}{4x + 5}\bigg]$

$\rm :\longmapsto\:y = \dfrac{1}{2} \bigg[\dfrac{4x + 5 + 1}{4x + 5}\bigg]$

$\rm :\longmapsto\:y = \dfrac{1}{2} \bigg[\dfrac{4x + 5}{4x + 5} + \dfrac{1}{4x + 5} \bigg]$

$\rm :\longmapsto\:y = \dfrac{1}{2} \bigg[1 + \dfrac{1}{4x + 5} \bigg]$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} \dfrac{1}{2} \bigg[1 + \dfrac{1}{4x + 5} \bigg]$

$\rm :\longmapsto\:y' =\dfrac{1}{2} \dfrac{d}{dx} \bigg[1 + \dfrac{1}{4x + 5} \bigg]$

We know,

$\boxed{ \bf{ \:\dfrac{d}{dx} \frac{1}{ {x}^{n} } = \frac{ - n}{ {x}^{n + 1} }}}$

and

$\boxed{ \bf{ \:\dfrac{d}{dx} k = 0}}$

So, using these results, we get

$\rm :\longmapsto\:y' =\dfrac{1}{2} \bigg[0 - \dfrac{1}{(4x + 5) {}^{2} }\dfrac{d}{dx}(4x + 5) \bigg]$

$\rm :\longmapsto\:y' =\dfrac{1}{2} \bigg[ - \dfrac{1}{(4x + 5) {}^{2} } \times 4 \bigg]$

$\rm :\longmapsto\:y' = - \: \dfrac{2}{ {(4x + 5)}^{2} }$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y' = - \: \dfrac{d}{dx} \: \dfrac{2}{ {(4x + 5)}^{2} }$

$\rm :\longmapsto\:y'' = ( - 2) \times \dfrac{( - 2)}{ {(4x + 5)}^{3} }\dfrac{d}{dx}(4x + 5)$

$\rm :\longmapsto\:y'' = \dfrac{4}{ {(4x + 5)}^{3} } \times 4$

$\rm :\longmapsto\:y'' = \dfrac{16}{ {(4x + 5)}^{3} }$

Hence,

$\rm :\longmapsto\:\boxed{ \: \: \bf{ \:y'' \: = \: \dfrac{16}{ {(4x + 5)}^{3} } \: \: }}$

Additional Information :-

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