please calculate dy/dx when y= [x^2(+2)]e^4x?

Question

That is : Y =[ (x squared) +2 ] * e^4x

Answer 1

Hi!

I would if I could, but I can't!

The kettle's just boiled and I have to make a brew!

Answer 2

y = [x² + 2]*e^(4x)

Using the product rule.
The first times the derivative of the second plus the second times the derivative of the first.

dy/dx = [x² + 2]*(4 e^(4x)) + e^(4x)*[2x]

factor to make the answer a little cleaner

=2e^(4x)(2[x² + 2]+x)

Answer 3

dy/dx when y=[x^2(+2)]e^4x = get a fricken life you nerd or even better try the opposite sex to spend your time on instead of a book .....dweeeeeeeeeeeeb

Answer 4

Use the product rule:
(f(x)g(x))'=f'(x)g(x)+f(x)g'(x).
You should get:
y' = 2xe^4x + 4(x^2 + 2)e^4x
= (4x^2 + 2x + 8)e^4x

Answer 5

Jimbo is right - use the product rule.

so (fg)' = f' g + f g'. so use f = x^2 + 2 and g = e^4x (and ' means differential)

so f' = 2x, and g' = 4e^4x

so (fg)' = (2x)(e^4x) + (x^2 +2)(4e^4x)

Expand the brackets etc and i'm sure it will tidy up nicely. write it on paper and double check evrything. good luck.

Answer 6

y= sqrt ((3x+a million)^3/(9x+2)) enable Y = (3x+a million)^3 / 9x+2 y=sqrt(Y) dY/dx = vdu - udv / v^2 = (9x+2)(3(3x+a million)^2) - (3x+a million)^3*9 / (9x+2)^2 = 3(3x+a million)^2 (9x+2 - (3x+a million)3) / (9x+2) ^2 = -3(3x+a million)^2 / (9x +2)^2 dy/dx = dy/dY * dY/dx dy/dY = a million/(2 sqrt(Y)) = a million/(2 sqrt ((3x+a million)^3 / 9x+2) * (-3(3x+a million)^2 / (9x +2)^2) there are too many indicators i'd have blended them up someplace, you should write down it on paper to confirm i have been given them maximum superb (no paper with me now)

Answer 7

You need to use the Product rule.

Answer 8

Wait till monday and ask your teacher. Or just take your anorak off.

Answer 9

y= [x^2(+2)]e^4x
y'=2x*e^(4x)+4e^(4x)(x^2+2)
=2x*e^(4x)+4x^2*e^(4x)
+8e^(4x)
=e^(4x)*(2x+4x^2+8)
=2e^(4x)*(2x^2+x+4)

i hope that this helps