Question

for the curve xy(x+y)=84 find dy/dx at (3,4)
b) find the gradient of the curve x^3-2y^3-3xy at the point (2,1)

Answer 1

\[x'*y*(y+x)+x*y'*(x+y)+x*y*((x+y))'\]

Answer 2

xy(x+y)=84
x^2y + xy^2 = 84
x^2y' + 2xy + 2xyy' + y^2 = 0
y'(x^2 + 2xy) = -(y^2 + 2xy)
y' = -(y^2 + 2xy) / (x^2 + 2xy)
Put x = 3, y = 4 and evaluate y'

Answer 3

@aum shouldnt we differentiate xy using product rule and then using sum rule?

Answer 4

for the inside brackets.

Answer 5

d/dx(x)*y(x+y)+dy/dx(y)*x*(x+y)+xy(d/dx(x)+dy/dx(y))

Answer 6

I just used algebraic expansion of xy(x+y) which makes it a bit simpler to differentiate.
xy(x+y) = x^2y + xy^2

Or you can keep xy(x+y) as is and differentiate.

Both should yield the same result.

Answer 7

Oh you're right thank you :) but can i ask one more question

Answer 8

Sure.

Answer 9

the curve y=e^x+4e^2x has one stationary point
find the x-coordinate of this point.
Determine whether the stationary point is maximum or minimum.

Answer 10

Find y', equate it to zero and solve for x.
You can either do a second derivative test or an increasing/ decreasing test to determine if it is a minimum or a maximum.

Answer 11

Is it \(\Large y = e^x + 4e^{2x}\) ?

Answer 12

or \(\Large y = e^x + 4e^{2}x\) ?

Answer 13

I don't think either one of those curves has a stationary point. Was there some typo in the function?

Answer 14

@aum
\[y=e^x+4e ^{-2x}\]
\[0=e^x+(4*e ^{-2x})\]\[e^x+(0*e ^{-2x}+4*e ^{-2x})\]\[-2*4*e ^{-2x}=-e^x\]\[-8e ^{-2x}=-e^x\]

Answer 15

Should i ln both sides or log both sides

Answer 16

This function makes sense. You left out the negative sign before 4e^(-2x)

Answer 17

am sorry :/

Answer 18

But no need to apply product rule when it is a constant.
d/dx (4 * e^(-2x)) = 4 * d/dx (e^(-2x)) = -8e^(-2x)

-8e^(-2x) = -e^x
8e^(-2x) = e^x
8e^(-2x) / e^x = 1
8e^(-2x - x) = 1
8e^(-3x) = 1
8 = e^(3x)
3x = ln(8)
x = ln(8) / 3

Answer 19

x = ln(8) / 3 = ln(2^3) / 3 = 3ln(2) / 3 = ln(2)

Answer 20

I dont get this part 8e^(-2x) = e^x
Did you divide by e^x

Answer 21

Isn't that what you had before after taking the derivative and equating it to zero? I started from that. Look at the last line in your earlier post.

Answer 22

oh now i get you divided by -e^x

Answer 23

yes.

Answer 24

thank you so much :D

Answer 25

You are welcome.

Answer 26

on the maximum and minimum part we take the second derivative as you said