Calculus AB

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Sorry but I've got a lot of problems that I don't understand.

1) Let f(x)= (3x-1)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)/(1-2x)

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 -

    #1 f' = (3x+2)e^x
    since e^x is always positive, just determine where 3x+2 is positive, etc.

    #2 remember the point-slope form of a line.

    a) y= x^2 + (e^x)/(x^2 +1) at the point x=3.
    y' = 2x + (x-1)^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)(x-3)

    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
    y-20 = 14(x-2)

    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
    y-1 = -(x-1)
    y = -x+2
    That line meets y = -x/3

    5) Not quite sure what you mean, but if
    y = 1/(1-2x)
    y' = 2/(1-2x)^2
    y" = 8/(1-2x)^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.

    (a) x^3 + y^3 = 3xy^2
    3x^2 + 3y^2 y' = 3y^2 + 6xy y'
    y'(3y^2-6xy) = 3y^2-3x^2
    y' = (x^2-y^2)/(y^2-2xy)

    (b) cos(xy^2) = y
    -sin(xy^2)(y^2+2xyy') = y'
    y' = -(y^2 sin(xy^2))/(1+2xy sin(xy^2))

    (a) 2x^2-3y^2 = 4
    4x-6yy' = 0
    y' = 2x/3y
    y" = (2*3y - 3x*3y')/9y^2
    = (6y-9x(2x/3y))/9y^2
    = 2(1-x^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 -

    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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