exponential differentiation.

**Tweety** · January 11th 2010, 06:53 AM

Find the coordinates of the turning point on the graph of $y = x\times e^{2x}$ and determine its
nature.

I used the product rule and got $e^{2x} +2x\times e^{2x}$

so to find the coordinates of the turning point I set that first derivative equal to 0.

$e^{2x} +2x\times e^{2x} = 0$

$e^{2x} (1+2x) = 0$

$e^{2x} = 0$ or $1+2x = 0$

$x = -\frac{1}{2}$

and to find the 'nature' of the turning points I have to take the second derivative. in which I dont get the correct answer, heres my working,

$\frac{dy}{dx} = e^{2x} + 2xe^{2x}$

the derivative of $e^{2x} = 2e^{2x}$
using the product rule on $2x e^{2x}$

$2 \times e^{2x} + 2e^{2x} \times 2x$

$\frac{d^{2}y}{dx^{2}} = 2e^{2x} + 2 \times e^{2x} + 2e^{2x} \times 2x$

= $4e^{2x} (1+x)$

but the my book says that the correct answer is $\frac{d^{2}y}{dx^{2}} = 2(2x+1)e^{2x} + e^{2x}.2$

**tonio** · January 11th 2010, 08:02 AM

Originally Posted by Tweety

Find the coordinates of the turning point on the graph of $y = x\times e^{2x}$ and determine its
nature.

I used the product rule and got $e^{2x} +2x\times e^{2x}$

so to find the coordinates of the turning point I set that first derivative equal to 0.

$e^{2x} +2x\times e^{2x} = 0$

$e^{2x} (1+2x) = 0$

$e^{2x} = 0$ or $1+2x = 0$

$x = -\frac{1}{2}$

and to find the 'nature' of the turning points I have to take the second derivative. in which I dont get the correct answer, heres my working,

$\frac{dy}{dx} = e^{2x} + 2xe^{2x}$

the derivative of $e^{2x} = 2e^{2x}$
using the product rule on $2x e^{2x}$

$2 \times e^{2x} + 2e^{2x} \times 2x$

$\frac{d^{2}y}{dx^{2}} = 2e^{2x} + 2 \times e^{2x} + 2e^{2x} \times 2x$

= $4e^{2x} (1+x)$

but the my book says that the correct answer is $\frac{d^{2}y}{dx^{2}} = 2(2x+1)e^{2x} + e^{2x}.2$

$\frac{d(2xe^{2x})}{dx} = 2e^{2x}+4xe^{2x}$

$\frac{d(e^{2x})}{dx}=2e^{2x}$

Sum the above and you get the 2nd derivative is $2(2x+2)e^{2x}=2(2x+1)e^{2x}+2e^{2x}$

Tonio

Ps. Oh, and PLEASE: do not use X to denote multiplication...this is not elementary school!

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