Calculus AB
posted by Annie .
Sorry but I've got a lot of problems that I don't understand.
1) Let f(x)= (3x1)e^x. For which value of x is the slope of the tangent line to f positive? Negative? Zero?
2) Find an equation of the tangent line to the oven curve at the specified point. Sketch the curve and the tangent line to check your answer.
a) y= x^2 + (e^x)/(x^2 +1) at the point x=3.
b) y= 2x(e^x) at the point x=0
3) Suppose that the tangent line to the graph of f at (2,f(2)) is y=3x+4. Find the tangent line to the graph of xf(x).
4) The line that is normal to the curve x^2 + 2xy  3y^2=0 at (1,1) intersects at what other point?
5) Evaluate (d^2)/(dx^2) * (1)/(12x)
6) Find dy/dx in terms of x and y.
a) x^3 +y^3=3xy^2
b) cos(xy^2)=y
7) Find ((d^2)y)/(dx^2) in terms of x and y.
a) 2x^2 3y^2=4
b) y+ siny=x
I know it's a lot but I just really don't understand how to do them.

Calculus AB 
Steve
#1 f' = (3x+2)e^x
since e^x is always positive, just determine where 3x+2 is positive, etc.
#2 remember the pointslope form of a line.
a) y= x^2 + (e^x)/(x^2 +1) at the point x=3.
y' = 2x + (x1)^2/(x+1)^2 e^x
y(3) = 9+e^3/10
y'(3) = 6+e^3/4
So, you want the line
y(9+e^3/10) = (6+e^3/4)(x3)
b) y= 2x(e^x) at the point x=0
y' = 2(x+1)e^x
y(0) = 0
y'(0) = 2
The tangent line is thus
y = 2x
3) d/dx (x*f) = f + xf'
f'(2) = 3 since that is the slope of the tangent line.
f(2) = 10 since the line meets it there.
at x=2, x*f has slope f(2) + 2*2 = f(2)+4 = 14
So the tangent line there is
y20 = 14(x2)
4)x^2 + 2xy  3y^2=0
This is just a pair of intersecting lines (a degenerate conic)
The lines are y=x and y = x/3
At (1,1) the normal line is
y1 = (x1)
y = x+2
That line meets y = x/3
(3,1)
5) Not quite sure what you mean, but if
y = 1/(12x)
y' = 2/(12x)^2
y" = 8/(12x)^3
For 6 and 7, just remember the chain rule and the product rule. When there are x's and y's, there will be y' floating around.
6)
(a) x^3 + y^3 = 3xy^2
3x^2 + 3y^2 y' = 3y^2 + 6xy y'
y'(3y^26xy) = 3y^23x^2
y' = (x^2y^2)/(y^22xy)
(b) cos(xy^2) = y
sin(xy^2)(y^2+2xyy') = y'
y' = (y^2 sin(xy^2))/(1+2xy sin(xy^2))
7)
(a) 2x^23y^2 = 4
4x6yy' = 0
y' = 2x/3y
y" = (2*3y  3x*3y')/9y^2
= (6y9x(2x/3y))/9y^2
= 2(1x^2)/3y
(b) y+siny = x
y' + cosy y' = 1
y' = 1/(1+cosy)
y" = siny/(1+cosy)^2 y'
= siny/(1+cosy)^3 
Calculus AB 
Annie
Thank you so much. I kind of understand the chain rule and product rule, but it's always hard for me to determine how to do/use them.
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