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2 years ago

If g(x)=x/e^x find g^(n)(x)

1 answer:

7 0

This equation is a very short question that has a very long answer,
Let me show you how to get the answer,
g(x) = x/e^x

One thing you can do is find the first few derivatives, and then you can validate the solution by induction. 

g'(x) = [ (1)e^x - x(e^x) ] / (e^x)^2 
= [e^x - xe^x] / [e^x]^2 
= (e^x)(1 - x) / (e^x)^2 
= (1 - x)/e^x 

g''(x) = [ (-1)(e^x) - (1 - x)(e^x) ] / (e^x)^2 
= [ -e^x - e^x + x e^x ] / (e^x)^2 
= [ -2e^x + x e^x ] / (e^x)^2 
= (e^x)(-2 + x) / (e^x)^2 
= (-2 + x) / (e^x) 
= (-1)(2 - x)/e^x 

g'''(x) = [ (1)(e^x) - (-2 + x)(e^x) ] / (e^x)^2 
g'''(x) = [ e^x - (-2e^x + x e^x) ] / (e^x)^2 
g'''(x) = [ e^x + 2e^x - x e^x ] / (e^x)^2 
g'''(x) = [ 3e^x - x e^x ] / (e^x)^2 
g'''(x) = (e^x)(3 - x) / (e^x)^2 
g'''(x) = (3 - x)/e^x 

By inspection, it appears the pattern is 

g^(n)(x) = (-1)^(n - 1) (n - x)/e^x

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Step-by-step explanation:

Here we have , A circle has diameter of 11 cm A square has side length of 7 cm . Use Pythagoras’ Theorem to show that the square will fit inside the circle without touching the edge of the circle . Let's find out:

We know the concept that for any square to fit inside the circle without touching the edge of circle , diagonal of square must be less than diameter of circle . Let's find out length of diagonal by using Pythagoras Theorem :

$Hypotenuse ^2 = Perpendicular^2+Base^2$

For a square , $Perpendicular = base = side$

⇒ $Diagonal^2 = 2(side)^2$

⇒ $Diagonal= \sqrt{2}(side)$

⇒ $Diagonal= \sqrt{2}(7)$

⇒ $Diagonal= 9.9cm$

Since length of diagonal ( $Diagonal= 9.9cm$ ) is less than diameter of circle ( 11 cm ) , Therefore , the square will fit inside the circle without ruching the edge of the circle.

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Step-by-step explanation:

After one year

A=p(1+r/n)^nt

=2000(1+0.03/12)^12*1

=2000(1+0.0025)^12

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=2000(1.0304)

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After two-years

A=p(1+r/n)^nt

=2060.8(1+0.03/12)^12*2

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After three years

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