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C3 query really bugging me

The curve has the equation, 'y= e^2x tanx'.
I am asked to show the turning points on this curve occur where
'tanx = 1'

I differentiated the equation and got:
2e^2x . tanx + e^2x sec^2 x = 0

then I took 'e^2x' out so that I have:

e^2x (2tanx + sec^2 x) = 0

afterwards, I changed 'sec^2 x' into '1+tan^2 x' so....

e^2x (2tanx + 1+tan^2 x) = 0

I sense that I need to factorise it, but how do the get rid of the
'e^2x' outside the brackets
Reply 1
Avatar for Pheylan
Pheylan
W.H.T
The curve has the equation, 'y= e^2x tanx'.
I am asked to show the turning points on this curve occur where
'tanx = 1'

this means that when you equate dy/dx to 0, you need to simplify it until you get tanx - 1 = 0

so you can just divide by the e^2x
e^2x=0

or
2 tanx+1+tan^2x=0

I think. Then factorise the tans.
W.H.T


e^2x (2tanx + 1+tan^2 x) = 0

I sense that I need to factorise it, but how do the get rid of the
'e^2x' outside the brackets


divide 0 by e^2x. Simples
Reply 4
Avatar for gdunne42
gdunne42
21
TylerDurden
e^2x=0

or
2 tanx+1+tan^2x=0

I think. Then factorise the tans.


noting of course that e^2x cannot equal 0
Reply 5
Avatar for maxfire
maxfire
14
Divide both sides by e2x
(tanx+1)2=0
Reply 6
Avatar for unamed
unamed
13
e2xe^{2x} can not equal 0.
Therefore, simply factorise tan2x+2tanx+1tan^2x+2tanx+1
and you would therefore get: (tanx+1)2(tanx+1)^2

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