#1
Hey, just a quick question to see if anyone out there can help with this quickly

If I differentiated Ln3x it gets split up to Ln3 + Lnx

Now Lnx goes to 1/x but Ln3 goes to 0; can someone bust out the reason why? My minds gone a bit rusty over the holidays
#4
well its been a couple of months since I did calculus, which means I've had enough time to forget most of it but ln(3) is a constant so it should disappear I think and dont quote me but ln(x) differentiates to 1/x. hope that helps, but I might be wrong
MaKing thE possiBlE...
...totaLlY impossible
#5
this is the easiest way to remember differentiation of log functions

when u differentiate them

Ln x = 1/x

the top is the differentiation of the original value and the bottom is the original value

so lnx^4 = 4x^3/x^4

understand? if you dont ill explain it better

Ln3 = 0/3 therefore = 0
Gear:
Maton MS503
Squier Strat
Randall RG75DG3+
H & K 100W Switchblade
Dunlop Original Crybaby
Digitech Crossroads
Digitech Jamman
Zoom G1

Quote by Sol9989
Caramello wins life.

Quote by A8039077
Caramello, that's mother****ing genius!
#6
Quote by Caramello Ruell
this is the easiest way to remember differentiation of log functions

when u differentiate them

Ln x = 1/x

the top is the differentiation of the original value and the bottom is the original value

so lnx^4 = 4x^3/x^4

understand? if you dont ill explain it better

Ln3 = 0/3 therefore = 0


While that gets the correct answer, it's not really correct.

If y=ln3, then dy/dx is the gradient of that graph for a given x value. Since that graph is a horizontal line, its gradient is 0 for all x, hence dy/dx = 0.