Please calculate dy/ dx Ln [ ( 2x^3 (+ 3x ) ] and of Ln ( x+ y ) = e^(x/y )?

Question

Ln ( 2x qubed) + 3x. '3x' is not in the exponent !

Answer 1

Would you be offended if I didn't?

Answer 2

1. dy/dx Ln (2x^3)+ 3x = (1/2x^3)*(6x^2) + 3
= (3/x) + 3
2. the second equation is an implicit equation and therefore you have to implicitly differentiate the function in order to find its dy/dx
Ln ( x+ y ) = e^(x/y )
[1/(x+y)]*1 + dy/dx [1/(x+y)]*1] = (1/y)*e^(x/y)+ (-x/y^2)*(dy/dx)
*e^(x/y )
(dy/dx)*[1/(x+y) + (x/y^2)*e^(x/y )]= (1/y)*e^(x/y) - 1/(x+y)
dy/dx = [(1/y)*e^(x/y) - 1/(x+y)] / )*[1/(x+y) + (x/y^2)*e^(x/y)]

Answer 3

Question 1
y = ln.(2x³ + 3x) and let u = (2x³ + 3x)
y = ln.u and du/dx = 6x² + 3
dy/du = 1/(u) = 1/(2x³ + 3x)
dy/dx = (dy/du).(du/dx)
dy/dx = 1/(2x³ + 3x).(6x² + 3)
dy/dx = 3.(2x² + 1) / x.(2x² + 3)

Question 2
ln(x + y) = e^(x/y)
e^(x + y) = e^(x/y)
x + y = x/y
xy + y² = x
Differentiate w.r.t. x:-
1.y + (dy/dx).x + 2y.(dy/dx) = 1
(x + 2y).(dy/dx) = 1 - y
dy/dx = (1 - y) / (x + 2y)

Answer 4

1)(dy/dx )Ln [2x^3 +3x ]
=(6x^2+3)*(1/(2x^3+3x)
=(6x^2+3)/(2x^3+3x)

2)here we use implicit
differentiation;
let fx' be a partial wrt x
let fy' be a partial wrt y

ln(x+y)-e^(x/y)=0
fx'=1/(x+y)-1/y*e^(x/y)
=(y-(x+y)e^(x/y)/(x+y)y
fy'= 1/(x+y)+(x/y^2)*e^(x/y)
=(y^2+x(x+y)*e^(x/y))
/(x+y)y^2

dy/dx= -fx'/fy'
= -y(y-(x+y)*e^(x/y))
/(y^2+x(x+y)*e^(x/y))
= (y(x+y)*e^(x/y)-y^2)
/(x(x+y)*e^(x/y)+y^2)

i hope that this helps

Answer 5

d/dx Ln[2x³ + 3x] = (6x²+3) / (2x³+3x)

Ln[x+y]=e^(x/y)
Differentiate both sides:
(1+dy/dx)/(x+y) = [(y+x dy/dx)e^(x/y)]/y²
Now rearrange to isolate dy/dx...

Answer 6

OMG! I used to love calculus at school, looking at that I don't know why! Good luck if somone actually does your homework for you!

Answer 7

at this time of night you have to be joking, try it in the morning.

Answer 8

Dont tell me what to do!!!!

Answer 9

what would be the point decimal or otherwise!!!

Answer 10

Tell you what, YOU do it.