y = ln(4x/5-3x)
What I have done is -
ln4x - ln(5-3x)
= dy/dx = 1/4x - 1/(5-3x)
= (5-3x-4x)/4x(5-3x)
Which is the wrong answer, please help?
I hope I have posted this in the right place, my apologies if not?
y = ln(4x/5-3x)
What I have done is -
ln4x - ln(5-3x)
= dy/dx = 1/4x - 1/(5-3x)
= (5-3x-4x)/4x(5-3x)
Which is the wrong answer, please help?
I hope I have posted this in the right place, my apologies if not?
U have to use the chain rule, because you're dealing with composite functions:
You're first step is good. But then it went wrong, because you didn't use the chain rule:
If you want to calculate for example:
$\displaystyle D[\ln(5-3x)]$
Let $\displaystyle 5-3x=u$
so you get:
$\displaystyle D[\ln(u)]=\frac{Du}{u}=\frac{D(5-3x)}{5-3x}= \frac{-3}{5-3x}$
Use also the chainrule to calculate $\displaystyle D[\ln(4x)]$ or notice that
$\displaystyle \ln(4x)=\ln(4)+\ln(x)$