dcgirl16 said:
the last part i don't know how they got any of it they got xlnx/e^x[1/x+(1/xlnx)-1]
=.="
Say you have an expression, of which derivatives is very "hard" to find using the normal way, i.e, using the
Quotient Rule, the
Chain Rule,
Product Rule, etc... then, the next thing you should try is to use logarithms, i.e, to simplify the problem.
Say, you want to find the derivative of:
y = \frac{x \ln x}{e ^ x}.
Of course, you can use a combination of the
Product Rule, and
Quotient Rule. But, well, it'll be
messy, and you may make mistake somewhere. So it's best to take logarithm of both sides, and change all to
sums (since, the
Sum Rule, and
Quotient Rule are far easier to use than those other rules).
We'll have:
\ln y = \ln \frac{x \ln x}{e ^ x}.
Apply some properties of logarithm here:
1. \ln \left( \frac{a}{b} \right) = \ln(a) - \ln (b)
2. \ln \left( ab \right) = \ln(a) + \ln (b)
3. \ln \left( a ^ \beta \right) = \beta \ln(a)Here we go:
\ln y = \ln \frac{x \ln x}{e ^ x} = \ln (x \ln (x)) - \ln (e ^ x) (using
(1)).
... = \ln (x) + \ln ( \ln (x)) - \ln (e ^ x) (using
(2)).
... = \ln (x) + \ln ( \ln (x)) - x \ln (e) = \ln (x) + \ln ( \ln (x)) - x (Using
(3)).
----------------
So, we have:
\ln y = \ln (x) + \ln ( \ln (x)) - x
If f(x) = g(x), then of course, we'll have f'(x) = g'(x), right? So, take the derivatives of both sides with respect to x, we have:
(\ln y)'_x = (\ln (x) + \ln ( \ln (x)) - x)'_x
\Rightarrow \frac{y'_x}{y} = (\ln (x))'_x + (\ln ( \ln (x)))'_x - (x)'_x
\Rightarrow \frac{y'_x}{y} = \frac{1}{x} + \frac{(\ln(x))'_x}{\ln(x)} - 1
\Rightarrow \frac{y'_x}{y} = \frac{1}{x} + \frac{\frac{1}{x}}{\ln(x)} - 1 = \frac{1}{x} + \frac{1}{x \ln(x)} - 1
Now, we have found \frac{y'_x}{y}, but well, what we need is y'
x, i.e the derivatives of y with respect to x, so let's multiply both sides by y to get:
... \Rightarrow y'_x = y \left( \frac{1}{x} + \frac{1}{x \ln(x)} - 1 \right)
But, what's
y? Well, it's y = \frac{x \ln x}{e ^ x}, our former function.
So, change y to x, we have:
... \Rightarrow y'_x = \left( \frac{x \ln x}{e ^ x} \right) \times \left( \frac{1}{x} + \frac{1}{x \ln(x)} - 1 \right)
Yeah, we've done it. :)
Is there anything unclear?
