Originally Posted by

**bhuang** I can't get a clear graph on here, but you can graph it on a graphing calculator, the graph of the curve with equation: y=e^(2x)cosx

On this graph, P(a,b) is labelled in the first quadrant as the graph is increasing.

a) Show that the first derivative=e^(2x)(2cosx-sinx)

b) Find second derivative

At the point P(a,b), the second derivative = 0.

c) Use results from a) and b) to prove that

i) tan *a* = 3/4

ii) the gradient of the curve at P is e^(2a)

Ok, so I can do a) and b), the second derivative is e^(2x)(3cosx-4sinx) and this is correct. I can prove c)i), but I dont know how to prove ii).

If I substitute 'a' into the equation of the derivative, then it's: e^(2a)(2cosa-sina), how is the answer e^(2a)???