don't understand at all...
find the slope of the tangent line to the curve y=xe^x at (1,e)
To find the slope of the tangent line to a curve at a specific point, you need to take the derivative of the equation that represents the curve. In this case, the curve is represented by the equation y=xe^x.
Step 1: Take the derivative of the curve equation.
To find the derivative of y=xe^x, you can use the product rule. The product rule states that if you have two functions u and v, then the derivative of their product is given by the formula d(uv)/dx = u*dv/dx + v*du/dx.
In this case, u = x and v = e^x. So, we can apply the product rule:
dy/dx = x * d(e^x)/dx + e^x * d(x)/dx
Now, let's find the derivatives of each part:
- The derivative of e^x with respect to x is simply e^x.
- The derivative of x with respect to x is 1.
Replacing these values, we have:
dy/dx = x * e^x + e^x * 1
= x*e^x + e^x
Step 2: Evaluate the derivative at the point (1,e).
To find the slope of the tangent line at a specific point, we need to substitute the values of x and y at that point into the derivative. In this case, we want to evaluate the derivative dy/dx at the point (1,e).
Substituting x = 1 and y = e into the derived equation, we have:
slope = dy/dx
= (1)*e^1 + e^1
= e + e
= 2e
Therefore, the slope of the tangent line to the curve y=xe^x at the point (1,e) is 2e.
What don't you understand.
This is very basic and fundamental Calculus, which is what you titled your post.
take derivative, that is your slope of the tangent.
Sub in x = 1 to get the numerical value of that slop
Use the method you learned in grade 8 or 9 to find the equation of the line, when you have the slope and a point on the line