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.
#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
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.
The partial derivative formula might help you hear!
No problem! I'll walk you through each question and explain how to solve them step by step.
1) To find the slope of the tangent line to the function f(x), we need to take the derivative of f(x) with respect to x and analyze its sign. Let's start by finding the derivative:
f(x) = (3x - 1)e^x
To take the derivative, we can use the product rule:
f'(x) = (3x - 1)e^x + (3)e^x
Simplifying, we get:
f'(x) = (3x + 2)e^x
Now, let's analyze the sign of the derivative to determine when the slope of the tangent line is positive, negative, or zero:
When f'(x) > 0, the slope of the tangent line is positive.
When f'(x) < 0, the slope of the tangent line is negative.
When f'(x) = 0, the slope of the tangent line is zero.
To find the value of x for which the slope is positive, we need to solve the inequality:
(3x + 2)e^x > 0
Similarly, for negative slope:
(3x + 2)e^x < 0
And for zero slope:
(3x + 2)e^x = 0
To solve these inequalities and equation, we need to consider the sign of e^x (which is always positive) and the sign of (3x + 2):
1. If (3x + 2) > 0, then the sign of (3x + 2)e^x is the same as the sign of e^x.
2. If (3x + 2) < 0, then the sign of (3x + 2)e^x is the opposite of the sign of e^x.
To find the values of x for each case, you can set up a sign chart or analyze intervals where (3x + 2) is positive or negative.
2) To find the equation of the tangent line, we can use the formula for the slope-intercept form of a line, y = mx + b, where m represents the slope of the tangent line and b represents the y-intercept.
a) Let's find the equation of the tangent line for the curve y = x^2 + (e^x)/(x^2 + 1) at the point x = 3.
First, take the derivative of the curve to find the slope of the tangent line:
f'(x) = 2x + [(2x)(x^2 + 1) - (e^x)(2x)] / (x^2 + 1)^2
Evaluate f'(x) at x = 3 to find the slope of the tangent line:
m = f'(3)
Once you have the slope, we can find the equation of the tangent line using the point-slope form:
y - y1 = m(x - x1)
Substitute the values of m, x1 = 3, and y1 = f(3) into the equation to get the equation of the tangent line.
For part b), follow the same steps using the given function y = 2x(e^x) and the specified point x = 0.
After finding the equation of the tangent line, sketching the curve and the tangent line can help you visually verify if the equation is correct.
3) To find the tangent line for the graph of xf(x), we need to find the derivative of xf(x) and use the given tangent line equation.
First, find the derivative of xf(x):
f'(x) = (x)(f'(x)) + f(x)
Given that the tangent line to the graph of f(x) at (2, f(2)) is y = 3x + 4, we know the slope of the tangent line is 3 and the point of tangency is (2, f(2)).
Now, substitute the value of f(2) into the equation y = 3x + 4 to find the y-coordinate of the point of tangency.
After finding the point (2, f(2)), substitute the slope and the coordinates into the point-slope form equation to find the equation of the tangent line for xf(x).
4) To find the other point of intersection for the normal line to the curve x^2 + 2xy - 3y^2 = 0 at (1, 1), we can use the fact that the product of the slopes of two perpendicular lines is -1.
First, find the slope of the tangent line to the curve at (1, 1).
Next, using the fact that the product of the slopes of the normal and tangent lines is -1, calculate the slope of the normal line.
Now, you have the slope of the normal line and the point of intersection (1, 1). Use the point-slope form equation to find the equation of the normal line.
To find the other point of intersection, you can substitute the equation of the normal line and solve for the coordinates.
5) To evaluate the expression (d^2/dx^2) * (1/(1-2x)), we need to find the second derivative of the expression.
First, find the derivative of the expression with respect to x using the product rule.
Then, differentiate again to find the second derivative.
The resulting expression will be the second derivative of the given expression.
6) To find dy/dx in terms of x and y, differentiate the given equations with respect to x using implicit differentiation.
a) For x^3 + y^3 = 3xy^2, differentiate both sides of the equation with respect to x and solve for dy/dx.
b) For cos(xy^2) = y, differentiate both sides of the equation with respect to x and solve for dy/dx.
7) To find ((d^2)y)/(dx^2) in terms of x and y, we need to differentiate the given equations twice with respect to x using implicit differentiation.
a) For 2x^2 - 3y^2 = 4, differentiate both sides with respect to x to find the first derivative. Then differentiate the first derivative with respect to x to find the second derivative.
b) For y + sin(y) = x, differentiate both sides with respect to x to find the first derivative. Then differentiate the first derivative with respect to x to find the second derivative.
I hope this explanation helps you understand how to approach each problem. If you have any further questions, feel free to ask!