Final answer:
To find the equation of the tangent to the curve y=e²ˣ² at the point (2,e⁸), we need to find the derivative of the function and substitute the x-value of the given point into the derivative. The derivative of y=e²ˣ² is 2e²ˣ², so the slope of the tangent line at (2,e⁸) is 2e²². Using the slope-point form, the equation of the tangent line is y - e⁸ = 2e²²(x - 2).
Step-by-step explanation:
To find the equation for the tangent to the curve y=e²ˣ² at the point (2,e⁸), we need to find the derivative of the function y=e²ˣ² and substitute the x-value of the given point into the derivative.
The derivative of y=e²ˣ² can be found using the chain rule: dy/dx = 2e²ˣ² * ln(e) * dx/dx which simplifies to dy/dx = 2e²ˣ²
Substituting x=2 into the derivative equation: dy/dx = 2e²² which gives us the slope of the tangent line at the point (2,e⁸). Now we can use the slope-point form of a line to find the equation of the tangent line: y - e⁸ = 2e²²(x - 2)