Question

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.

Answer 1

#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
(3,-1)

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.

6)
(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))

7)
(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

Answer 2

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.

Answer 3

The partial derivative formula might help you hear!