find the slope of the tangent line at the point P(1,1) on the graph of e^x-y=2x^2-y^2
What is the exponent of e: x or x-y?
1,1 is a point on the curve of
e^(x-y) = 2x^2 -y^2
so I will use that.
Differentiate both sides implicitly with respect to x (treating y as an implicit function of x) and solve for the value of dy/dx when x = 1.
e^(x-y)*(1 - dy/dx) = 4x - 2y*dy/dx
So at x = 1, y = 1,
1 - dy/dx = 4 - 2 dy/dx
dy/dx = 3
To find the slope of the tangent line at the point P(1,1) on the graph of e^x - y = 2x^2 - y^2, we will differentiate the equation implicitly and substitute the values of x and y from point P(1,1).
Let's start by differentiating both sides of the equation with respect to x:
d/dx (e^x - y) = d/dx (2x^2 - y^2)
Using the chain rule, the derivative of e^x with respect to x is e^x, and the derivative of -y with respect to x is -dy/dx. Similarly, the derivative of -y^2 with respect to x is -2y * (dy/dx). Applying the chain rule again, the derivative of 2x^2 with respect to x is 4x.
Therefore, we have:
e^x - dy/dx = 4x - 2y * (dy/dx)
Next, let's substitute the values of x and y from point P(1,1):
e^1 - dy/dx = 4(1) - 2(1) * (dy/dx)
e - dy/dx = 4 - 2(dy/dx)
To find dy/dx, let's isolate it:
dy/dx - 2(dy/dx) = 4 - e
- dy/dx = 4 - e
dy/dx = e - 4
So, the slope of the tangent line at the point P(1,1) is e - 4.
To find the slope of the tangent line at the point P(1,1) on the graph of the equation e^x - y = 2x^2 - y^2, we need to find the derivative of the equation with respect to x. The derivative will give us the slope of the tangent line at any given point.
Let's start by rearranging the equation to isolate y:
e^x - 2x^2 = y + y^2
Now, we can find the derivative of both sides with respect to x.
The derivative of e^x with respect to x is simply e^x.
The derivative of 2x^2 with respect to x is 4x (using the power rule).
The derivative of y with respect to x is dy/dx, which represents the slope of the tangent line.
The derivative of y^2 with respect to x can be found using the chain rule: (2y)(dy/dx).
Now, we can rewrite the equation with the derivatives:
e^x - 2x^2 = dy/dx + 2y(dy/dx)
Now, let's substitute the coordinates of the point P(1,1) into the equation.
e^1 - 2(1)^2 = dy/dx + 2(1)(dy/dx)
e - 2 = (dy/dx) + 2(dy/dx)
To solve for dy/dx, we need to combine like terms:
(dy/dx) + 2(dy/dx) = e - 2
3(dy/dx) = e - 2
dy/dx = (e - 2) / 3
Therefore, the slope of the tangent line at the point P(1,1) on the given graph is (e - 2) / 3.