Ive managed to do the first 2 questions
Any help on the last 2 would be appreciated
Hey i am having a bit of trouble with these differentiation problems;
1. Find the equation to the tangent to y=e^-x, at the point where x = 1
2. Find the equation to the tangent to y= ln(2-x) at the point where x = -1
3. The tangent at x = 1 to y = x^2e^x cuts the x and y axis at A and B respectively. Find the coordinate of A and B
4. Find the equation of the normal to y = ln √ x at the point where y = -1
Im not sure about any of these but for the first 2 i think you have to differentiate first but im nt really sure..
thanks!
Q.1 : y=e^-x equation of tangent at point of abcissa x = 1
1. y'=-e^-x
2. y'(1)=-e
3. y(1)=e^-1
then the equation of the tangent is : Yt=y'(1) ( X-1 ) + y(1)
Yt=-e(x-1)+e^-1 ; Yt=-ex+(e+e^-1).
Q.2 : y= ln(2-x) x=-1
1. y'= ( -1 ) / ( 2-x )
2. y'(-1) = -1/3
3. y(-1) = ln(3)
Yt=y'(-1) ( X+1 ) + y(-1)
Yt= (-1/3) ( X+1 ) + ln(3)
Yt= (-1/3)X+(ln(3) -1/3 )
Q.3
first you get the equation of the tangent just as Q1 and Q2 the general form of the equation of the tangent Yt to a point of absissa Xt on a graph of equation Y is : Yt = Y'(Xt) ( X-Xt ) + Y(Xt)
1. the equation of the tangent here is : Yt=(3e)X-2e
2. the point A is where Yt=0 , => A(2/3,0)
3. the point B is where X=0 , ==> B(0,-2e)
Q.4 i didnt undestand what do u mean by the normal to y=ets ...
You need the coordinates of the point you're finding the normal at. You're given the y-coordinate. To find the x-coordinate, solve for x.
To get the gradient, recall that so you first need to find at the value of x found above.
Now you have a point and a gradient so you should be able to get the equation of the normal.