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Differentiation help please...

A curve has equation y= e4xe^{-4x}(x2x^{2}+2x-2)
Show that dydx\frac{dy}{dx} = 2e4x2e^{-4x}(55-3x-2x22x^{2})

How do you do this? I differentiated the e4xe^{-4x} and got 4ex4x-4ex^{-4x} and the differentiated the bracket bit to get (2x+2) :/
(edited 13 years ago)
Original post by Nkhan


A curve has equation: e4xe^{-4x}(x2x^{2}+2x-2)
Show that dydx\frac{dy}{dx} = 2e4x2e^{-4x}(5-3x-2x22x^{2})



Well have you tried using the product rule?

(u x dv/dx) + (v x du/dx)

where u = the first part of your function and v = the second part of the function
(edited 13 years ago)
Reply 2
Original post by Mr_Muffin_Man
Well have you tried using the product rule?

(u x dv/dx) + (v x du/dx)

where u = the first part of your function and v = the second part of the function

I differentiated the e4xe^{-4x} and got 4ex4x-4ex^{-4x} and the differentiated the bracket bit to get (2x+2) :/
Reply 3
Original post by Nkhan
I differentiated the e4xe^{-4x} and got 4ex4x-4ex^{-4x} and the differentiated the bracket bit to get (2x+2) :/


Your function is product of an exponential and a polinomial function,
let u=f(x) and v=g(x) so y=f(x)*g(x)
Derivating y (dy/dx)you should use the product rule for these type of functions.
This rule: Y' =f'(x)*g(x) + f(x)*g'(x)
As others have said, product rule that shizzle.
Seriously? Apply the Product Rule!!!
Reply 6
[br][br]y = (e4x)(x2+2x2)[br][br]u=e4x[br]dudx =4e4x[br][br]v=x2+2x2[br]dvdx =2x+2[br][br]ProductRule=udv/dx+vdu/dx[br][br]=(e4x)(2x+2)4e4x(x2+2x2)[br][br]=2(e4x)(x+1)4e4x(x2+2x2)[br][br]=2(e4x)(x+1)2(e4x)2(x2+2x2)[br][br]=2e4x((x+1)2(x2+2x2))[br][br]=2e4x(53x2x2)[br][br][br][br]y\ =\ (e^-4x )(x^2 +2x-2)[br][br]u= e^-4x[br]\dfrac{du}{dx}\ = -4e^-4x[br][br]v= x^2 +2x-2[br]\dfrac{dv}{dx}\ = 2x+2[br][br]Product Rule = u dv/dx + v du/dx[br][br]= (e^-4x)(2x+2) - 4e^-4x (x^2 +2x-2)[br][br]= 2(e^-4x)(x+1) - 4e^-4x (x^2 +2x-2)[br][br]= 2(e^-4x)(x+1) - 2(e^-4x) 2(x^2 +2x-2)[br][br]= 2e^-4x ((x+1) - 2(x^2 +2x-2))[br][br]= 2e^-4x (5-3x-2x^2 )[br][br]

Hope that helps!q
(edited 13 years ago)
Reply 7
See more solved problems on product rule at http://www.math24.net/derivative-of-product-and-quotient.html.
Original post by Nkhan
A curve has equation y= e4xe^{-4x}(x2x^{2}+2x-2)
Show that dydx\frac{dy}{dx} = 2e4x2e^{-4x}(55-3x-2x22x^{2})

How do you do this? I differentiated the e4xe^{-4x} and got 4ex4x-4ex^{-4x} and the differentiated the bracket bit to get (2x+2) :/
product rule: dy/dx of uv = v(du/dx) + u(dv/dx)

uv = [e^(-4x)] (x^2+2x-2)

therefore u = [e^(-4x)] and v = (x^2 + 2x - 2)
du/dx = -4e^(-4x) and dv/dx = 2x + 2

using that formula:

(x^2 + 2x - 2)[-4e^(-4x)] + [e^(-4x)](2x+2)

simplifying will hopefully give you the 'show that' answer

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