Math, asked by sanjananewar9, 5 months ago

y= (1+x)/logx find out in double differenciation

Answers

Answered by Anonymous
37

Answer :

The second derivative of \sf{y = \dfrac{1 + x}{log(x)}} is \sf{-\dfrac{1}{(x)log(x)^{2}} + \dfrac{log(x) + 2x + 2}{\big\{(x^{2})log^{3}(x)\big\}}} \\ \\

Explanation :

To find :

The derivative of :

\sf{y = \dfrac{1 + x}{log(x)}}

Knowledge required :

  • Quotient rule of differentiation :

\sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{v\dfrac{d(u)}{dx} - u\dfrac{d(v)}{dx}}{v^{2}}}

  • Product rule of differentiation :

\sf{\dfrac{d(vu)}{dx} = u\dfrac{d(v)}{dx} + v\dfrac{d(u)}{dx}}

  • Derivative rule for log x :

\sf{\dfrac{d(log(x))}{dx} = \dfrac{1}{x}}

  • Derivative of a constant term is 0.

\sf{\dfrac{d(k)}{dx} = 0}

  • Exponent rule of differentiation :

\sf{\dfrac{d(x^{n})}{dx} = nx^{(n - 1)}}

Solution :

By using the quotient rule of differentiation and substituting the values in it, we get : [Here, u = (x + 1) and v = log(x)]

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x + 1}{log(x)}\bigg) = \dfrac{\big(log(x)\big)\dfrac{d(x + 1)}{dx} - (x + 1)\dfrac{\big(log(x)\big)}{dx}}{log^{2}(x)}} \\ \\

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x + 1}{log(x)}\bigg) = \dfrac{\big(log(x)\big)\dfrac{d(x)}{dx} + \dfrac{d(1)}{dx} - (x + 1) \times \dfrac{1}{x}}{log^{2}(x)}} \\ \\

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x + 1}{log(x)}\bigg) = \dfrac{\big(log(x)\big)}{log^{2}x} - \dfrac{\dfrac{(x + 1)}{x}}{log^{2}(x)}} \\ \\

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x + 1}{log(x)}\bigg) = \dfrac{1}{log(x)} - \dfrac{(x + 1)}{(x)log^{2}(x)}} \\ \\

\boxed{\therefore \sf{\dfrac{d}{dx}\bigg(\dfrac{x + 1}{log(x)}\bigg) = \dfrac{1}{log(x)} - \dfrac{(x + 1)}{(x)log^{2}(x)}}} \\ \\

Now let's find out the second Derivative of the equation :

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x + 1}{log(x)}\bigg) = \dfrac{1}{log(x)} - \dfrac{(x + 1)}{(x)log^{2}(x)}} \\ \\

From there,

First let us differentiate the terms :

Differentiation of \sf{\dfrac{1}{log(x)}} \\ \\ :

:\implies \sf{\dfrac{dy}{dx} = \dfrac{1}{log(x)}} \\ \\

:\implies \sf{\dfrac{dy}{dx} = log(x)^{-1}} \\ \\

By applying the chain rule of differentiation, we get :

:\implies \sf{\dfrac{dy}{dx} = \dfrac{d(log(x)^{-1})}{d(log(x))} \times \dfrac{d(log(x))}{dx}} \\ \\

:\implies \sf{\dfrac{dy}{dx} = (-1) \times log(x)^{(-1) - (-1)} \times \dfrac{1}{x}} \\ \\

:\implies \sf{\dfrac{dy}{dx} = -log(x)^{(-2)} \times \dfrac{1}{x}} \\ \\

:\implies \sf{\dfrac{dy}{dx} = -\dfrac{1}{log(x)^{2}} \times \dfrac{1}{x}} \\ \\

:\implies \sf{\dfrac{dy}{dx} = -\dfrac{1}{(x)log(x)^{2}}} \\ \\

\boxed{\therefore \sf{\dfrac{d}{dx}\bigg(\dfrac{1}{log(x)}\bigg) = -\dfrac{1}{(x)log(x)^{2}}}} \\ \\

Differentiation of \sf{\dfrac{(x + 1)}{(x)log^{2}(x)}} \\ \\

By using the quotient rule of differentiation and substituting the values in it, we get :

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\big\{(x)log^{2}(x)\big\}\dfrac{d(x + 1)}{dx} - (x + 1)\dfrac{d\big((x)log^{2}(x)\big)}{dx}}{\big\{(x)log^{2}(x)\big\}^{2}}} \\ \\

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\big\{(x)log^{2}(x)\big\}(1) - (x + 1)\dfrac{d\big((x)log^{2}(x)\big)}{dx}}{\big\{(x)log^{2}(x)\big\}^{2}}} \\ \\

\boxed{\begin{minipage}{8 cm}$\sf{Differentiation\:of\:(x)log^{2}(x)} \\ \\ :\implies \sf{\dfrac{d(vu)}{dx} = \dfrac{d((x)log^{2}(x)}{dx}} \\ \\ Now by using the product rule of differentiation, we get : \\ \\ :\implies \sf{\dfrac{d(vu)}{dx} = u\dfrac{d(v)}{dx} + v\dfrac{d(u)}{dx}} \\ \\ :\implies \sf{\dfrac{d(vu)}{dx} = x\dfrac{d(log^{2}(x))}{dx} + (log^{2}(x))\dfrac{d(x)}{dx}} \\ \\ :\implies \sf{\dfrac{d(vu)}{dx} = 2log(x) + log^{2}(x)} \\ \\ \boxed{\therefore \sf{\dfrac{d(vu)}{dx} = 2log(x) + log^{2}(x)}}$\end{minipage}}

Now by substituting the derivative of log²x , we get :

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\big\{(x)log^{2}(x)\big\}(1) - (x + 1)\dfrac{d\big((x)log^{2}(x)\big)}{dx}}{\big\{(x)log^{2}(x)\big\}^{2}}} \\ \\

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\big\{(x)log^{2}(x)\big\} - (x + 1)(2log(x) + log^{2}(x))}{\big\{(x^{2})log^{4}(x)\big\}}} \\ \\

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = -\dfrac{log(x) + 2x + 2}{\big\{(x^{2})log^{3}(x)\big\}}} \\ \\

\boxed{\therefore \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = -\dfrac{log(x) + 2x + 2}{\big\{(x^{2})log^{3}(x)\big\}}}} \\ \\

Now by substituting the values in the equation, we get :

:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x + 1}{log(x)}\bigg) =  -\dfrac{1}{(x)log(x)^{2}} - \bigg[-\dfrac{log(x) + 2x + 2}{\big\{(x^{2})log^{3}(x)\big\}}\bigg]} \\ \\

\boxed{\therefore \sf{\dfrac{d}{dx}\bigg(\dfrac{x + 1}{log(x)}\bigg) =  -\dfrac{1}{(x)log(x)^{2}} + \bigg[\dfrac{log(x) + 2x + 2}{\big\{(x^{2})log^{3}(x)\big\}}\bigg]}} \\ \\

Answered by riddhima8
2

Answer:

The derivative of \sf{y = \dfrac{1 + x}{log(x)}}y=

log(x)

1+x

is \begin{gathered}\sf{\dfrac{1}{log(x)} - \dfrac{(x + 1)}{(x)log^{2}(x)}} \\ \\\end{gathered}

log(x)

1

(x)log

2

(x)

(x+1)

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