Expert: Alon Mandes - 11/15/2008

Question
I need help for this a) differentiate (e^x)/(1+e^(-x)), b)find the equation of the tangent to the curve y=e^(3x-5) at the point where x=2, c)a particle moves in a straight line so that after t seconds its distance from a fixed point O is s metres,where s=t^2e^(2-t). Find the distance of the particle from O when it first comes to rest & its acceleration at that point. d)find the values of x for which the function (x^2-2x-1)e^(2x) has maximum or minimum values,distinguishing between them. I have tried them all but i'm stuck. Much appreciated.

Answer
a)In this suction we will use 2 rules :
1. (e^x)'=e^x , (e^-x)'=-e^-x
2. (f/g)'=(f'g-g'f)/g^2
So, (e^x)/(1+e^-x)'=[ (e^x)*(1+e^-x)-(e^x)*(-e^-x) ] / (1+e^-x)^2
=[e^x+1+1]/(1+e^-x)^2=(e^x+2)/(1+e^-x)^2

b)The tangent line of function y=f(x) is from the form ax+b
where a is y'(x) in point x=1. 1st let's find y(1):
y(1)=e^(3*1-5)=e^-2. Now let's derive y :
y'=[e^(3x-5)]'=3e^(3x-5). y'(1)=3e^-2. & this will be a.
Now we need to find b, to do so we substitute the point (1,e^-2)
in the tangent line : e^-2=a*1+b -> b=e^-2-a=e^-2 -3e^-2 , thus
Tangent Line :  (3e^-2)x+e^-2 -3e^-2=(3e^-2)(x+1)+e^-2.

d)In this section we will use the rule : (fg)'=f'g+fg', so
f(x)=(x^2-2x-1)e^2x -> f'(x)=(x^2-2x-1)'(e^2x)+(x^2-2x-1)(e^2x)'
f'(x)=(2x-2)(e^2x)+(x^2-2x-1)2e^2x=
f'(x)=2x[e^2x]-2[e^2x]+2x^2[e^2x]-4x[e^2x]-2[e^2x]
f'(x)=-2xe^2x-4e^2x+2x^2e^2x
f'(x)=e^2x[2x^2-2x-4]=(x-2)(x+1)e^2x
To find min/max we will set the derivative equal zero !
f'(x)=0 -> (x-2)(x+1)e^2x=0 -> (x-2)(x+1)=0 -> x=2 & x=-1.
To find which gives max & whic gives min , we must calculate 2nd
derivative : f''(x)={ e^2x[2x^2-2x-4] }'=
2e^2x*(4x-2)+e^2x*(2x^2-2x-4)=e^2x[2(4x-2)+2x^2-2x-4]=
e^2x[2x^2+6x-8]. Now we will chick :
f''(2)=e^4*[8+12-8] > 0 So x=2 is max
f''(-1)=e^-2*[2-6-8] < 0 So x=-1 is min

c)S(t)=t^2*e(2-t), the particle is at rest when the velocity is
zero ! ^ we know that the velocity is S'(t), so :
S'(t)=2te^(2-t)-t^2e^(2-t) , S'(t)=0 -> 2te^(2-t)-t^2e^(2-t) ->
2e^(2-t)-te^(2-t) //we divided by t
2=t //we divided by e^(2-t)
& the time when the particle rest is t=2.
S(2)=2^2*e(2-2)=4.
To find acceleration we derive 2nd time :
S''(t)=[ 2te^(2-t)-t^2e^(2-t) ]'
=2e^(2-t)-2te^(2-t)-2te^(2-t)+t^2e^(2-t)
S''(2)=2e^(2-2)-2*2e^(2-2)-2*2e^(2-2)+2^2e^(2-2)
=2-4-4+4=-2.

Alon.