find the tangent to the curve y=e^(x^2+3x)+2x at the point x=1
To find the tangent to the curve at the point x=1, we first need to find the slope of the curve at that point.
The slope of the tangent line to a curve is given by the derivative of the function at that point.
Given that the function is y=e^(x^2+3x)+2x, we first find the derivative of the function using the chain rule and product rule:
dy/dx = d/dx(e^(x^2+3x)) + d/dx(2x) = (2x + 3)e^(x^2+3x) + 2
Now, we can find the slope of the tangent line at x=1 by plugging in x=1 into the derivative:
dy/dx = (2(1) + 3)e^(1^2 + 3(1)) + 2 = (2 + 3)e^4 + 2 = 5e^4 + 2
So, the slope of the tangent line at x=1 is 5e^4 + 2.
Now, we need to find the point on the curve at x=1:
y(1) = e^(1^2 + 3(1)) + 2(1) = e^4 + 2
Therefore, the point on the curve at x=1 is (1, e^4 + 2).
Finally, we can write the equation of the tangent line in point-slope form:
(y - y1) = m(x - x1)
(y - (e^4 + 2)) = (5e^4 + 2)(x - 1)
y - e^4 - 2 = 5e^4x + 2x - 5e^4 - 2
y = 5e^4x + 2x
Therefore, the equation of the tangent line to the curve y=e^(x^2+3x)+2x at the point x=1 is y = 5e^4x + 2x.