Exponential Derivatives?

Question

"Find the tangent to the curve y=e^(2x) at the point where x=1." Do I need to know a special rule to do this question?

Answer 1

y=e^(2x)
find dy/dx = 1/2 *(e^(2x))
when x=1
dy/dx = 1/2 *(e^(2(1)))
then find -dx/dy. That is reverse dy/dx and add a negative sign. Then replace x=1 into y=e^(2x) to find the value of y
Equation of tangent
(Y - y) = -dx/dy(X - x)

Now calculate. Best Luck!!!

Answer 2

dy/dx = f ` (x) = 2.e^(2x)
f ` (1) = 2 = m
Tangent passes thro` (1 , e²)
y - b = m.(x - a)
y - e² = 2.(x - 1)
y = 2x - 2 + e²
y = 2x + 5.33