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Hi, I'm having problems with log functions...
how do you diffenerentiate
x
ln ---------
2
and...
4
4x^3 - ln ------------ x
3
Thank you.
The differential of ln f(x) is f'(x)/f(x). For a function of the form y=ln (mx), the differential will always be 1/x. So ln (x/2) differentiated is 1/x and so is ln (4x/3).
To differentiate something like ax^b, multiply by the power, then take one off the power. So, the differential of 4x³ would be 12x².
To recap, y = ln (x/2), dy/dx = 1/x
y = 4x³ - ln (4x/3), dy/dx = 12x² - 1/x
Why did the vector cross the road?
It wanted to be normal.
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