Could really use a lot of help here
what is the slope of the tangent line to
cos(x-y)=xe^x at point (0,3pi/2)
how so? It's just a straightforward implicit differentiation problem. (with a little chain rule and product rule thrown in)
-sin(x-y)(1 - y') = e^x + xe^x
1-y' = e^x(x+1)/-sin(x-y)
y' = 1 + e^x(x+1)/sin(x-y)
So, at (0,3pi/2),
y' = 1 + 1(0+1)/sin(0-3pi/2)
= 1 + 1/1
= 2
Oh, I see you're in need of some mathematical assistance! Well, fear not, for I, Clown Bot, am here to help!
To find the slope of the tangent line to the curve represented by the equation cos(x - y) = xe^x at the point (0, 3π/2), we need to apply some derivative fun! First, let's differentiate both sides of the equation with respect to x.
-d/dx(sinx - y) = d/dx(xe^x)
Now, let's simplify this expression a bit. The derivative of sin(x) with respect to x is just cos(x), and the derivative of y with respect to x is dy/dx. Note that x and y are related variables, so consider y as a function of x.
-cos(x - y) = xe^x + e^x(dy/dx)
Since we're interested in finding the slope at the point (0, 3π/2), we can substitute x = 0 and y = 3π/2 into our equation.
-cos(0 - 3π/2) = 0e^0 + e^0(dy/dx)
Simplifying further, we have cos(3π/2) = dy/dx.
Now, what's the slope when the argument of cosine is 3π/2? It's zero! Thus, the slope of the tangent line to the curve at the point (0, 3π/2) is zero.
Hope that helps, and remember, math can sometimes be a real juggling act!
To find the slope of the tangent line to the curve defined by the equation cos(x-y) = xe^x at the point (0, 3pi/2), we can follow these steps:
Step 1: Differentiate the equation with respect to x.
Step 2: Plug in the x and y values of the given point into the derivative.
Step 3: Solve for the slope.
Let's go through each step in detail:
Step 1: Differentiate the equation with respect to x.
Differentiating both sides of the equation cos(x-y) = xe^x with respect to x using the chain rule, we get:
[-sin(x-y)] * (1 - dy/dx) = xe^x + e^x * (1 - dy/dx)
Step 2: Plug in the x and y values of the given point into the derivative.
Substituting x = 0 and y = 3pi/2 into the derivative, we have:
[-sin(0 - 3pi/2)] * (1 - dy/dx) = (0)(e^0) + e^0 * (1 - dy/dx)
Simplifying further:
sin(3pi/2) * (1 - dy/dx) = (1 - dy/dx)
Step 3: Solve for the slope.
Now, we need to solve the equation sin(3pi/2) * (1 - dy/dx) = (1 - dy/dx) for dy/dx, which represents the slope of the tangent line:
[(1 - dy/dx) / (1 - dy/dx)] * sin(3pi/2) = (1 - dy/dx) / (1 - dy/dx)
sin(3pi/2) = 1
Since sin(3pi/2) equals 1, we have:
1 = 1 - dy/dx
Simplifying further:
dy/dx = 0
So, the slope of the tangent line to the curve cos(x-y) = xe^x at the point (0, 3pi/2) is 0.
Sure! I can help you with that. To find the slope of the tangent line to the curve, we need to find the derivative of the function cos(x-y) = x * e^x and then evaluate it at the given point (0, 3π/2).
To find the derivative, we can use the chain rule. Let's start by differentiating each term separately:
d(cos(x-y))/dx = -sin(x-y) * (d(x-y)/dx) = -sin(x-y) * (1 - 0) = -sin(x-y)
d(x * e^x)/dx = (d(x)/dx) * e^x + x * (d(e^x)/dx) = e^x + x * e^x = (1 + x) * e^x
Now, we can find the derivative of the entire equation using the chain rule:
d(cos(x-y))/dx = d(x * e^x)/dx
-sin(x-y) = (1 + x) * e^x
Next, we need to evaluate this derivative at the point (0, 3π/2). Since we need to find the slope of the tangent line, we are looking for the value of dy/dx at that point.
Plugging in x = 0 and y = 3π/2 into the derivative equation, we get:
-sin(0 - 3π/2) = (1 + 0) * e^0
-sin(-3π/2) = 1
Now, we have the value of dy/dx at the given point, which is 1. Therefore, the slope of the tangent line to the curve cos(x-y) = x * e^x at the point (0, 3π/2) is 1.