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
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Ln3 is a constant so it doesn't affect the gradient.
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
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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
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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.
Ah thank you to everyone who helped, made it much clearer, thank you