Differentiate y= x / ( 2+3ln(x) ) ?

Question

Can someone please show me the method for differentiating the following equation:

y= x / ( 2+3ln(x) )

Cheers

Answer 1

y= x / (2 + 3ln(x))

Simply use the quotient rule.

y' = [(2 + 3ln(x)) - x(3/x)] / (2 + 3ln(x))^2

y' = [2 + 3ln(x) - 3] / (2 + 3ln(x))^2

y' = [3ln(x) - 1] / (2 + 3ln(x))^2

Answer 2

The above can be thought of as the division between two functions; let's say f(x) = x and g(x) = 2+3ln(x)

Let's name the first derivatives of f(x) and g(x) as f'(x) and g'(x)
So we have:
f'(x) = 1
g'(x) = 3/x

the formula when we have two dividing functions is:
[f'(x)g(x)-f(x)g'(x)] divided by [(g(x)^2]

So we have:
[(2+3ln(x))-(x*3/x)] divided by [(2+3ln(x))^2]

so solve for the above equation:
[3ln(x)-1] divided by [(2+3ln(x))^2]

Answer 3

You need to use the quotient rule

d/dx (u/v) = (u'v - uv') / v^2

Where we let u = x and v = 3ln(x) + 2

so we get

[ (3ln(x)+2) d/dx(x) - d/dx (3ln(x)+2) ] / (3ln(x)+2)^2

The derivative of a sum is the sum of the derivatives

That is'

[ (3ln(x)+2) d/dx(x) - d/dx(2) + d/dx(3ln(x)) ] / (3ln(x)+2)^2

Now the derivative of a constant times a function is the same constant times the derivative of the funcion.

[ (3ln(x)+2) d/dx(x) - x - 3x d/dx(ln(x)) ] / (3ln(x)+2)^2

The derivative of ln(x) = 1/x

[ (3ln(x)+2) d/dx(x) d/dx (x) - 3] / (3ln(x)+2)^2

= [3ln(x) - 1 ] / (3ln(x)+2)^2

The end.

Answer 4

We know the quotient rule :
d/dx [f(x) / g(x)] = (g(x)f'(x) - f(x)g'(x))/([g(x)]^2)

f(x) = x
g(x) = 2 + 3ln(x)

g(x)f`(x) - f(x)g`(x) = 2 + 3ln(x) - x(3/x) = 3ln(x) - 1

Answer is (3ln(x) - 1)/[(2+3ln(x))^2]

Answer 5

Let u=x and v = 2 + 3 ln(x)

Therefore y = u/v,

Using the quotient rule, dy/dx = (V.du/dx -U.dv/dx)/V^2

du/dx = 1; dv/dx = 3/x

Therefore dy/dx = ((2+3 ln(x)).1-x.(3/x))/(2+3ln(x))^2
= (3 ln(x)-1)/(2+3ln(x))^2

Answer 6

y = u / v where u = x and v = 2 + 3lnx
du/dx = 1 and dv/dx = 3/x
By quotient rule:-
dy/dx= [(2 + 3lnx).1 - x.(3/x)] / (2 + 3lnx)²
= (3.lnx - 1) / (2 + 3lnx)²

Answer 7

y= x / ( 2+3ln(x) )

let u=1/(2+3lnx)
then,
du/dx=(-3/x)/(2+3ln(x))^2
let v=x,then dv/dx=1
dy/dx=u*dv/dx+v*du/dx
=1/(2+3lnx)*1+
(-3x/x)/(2+3ln(x))^2
=1/(2+3lnx)-3/(2+3lnx)^2
={(2+3lnx)-3}/(2+3lnx)^2
=(3lnx-1)/(2+3lnx)^2

i hope that this helps

Answer 8

Answer is:-
(3logx-1)/(2+3logx)(2+3logx)

Answer 9

dy/dx of ln(x) is x**-1.