$ \frac{\partial f}{\partial x}\dfrac{1}{(1-x)^2} = \dfrac{0 *(1-x)^2 - (-1)(1)}{(1-x)^2*(1-x)} $
But:
$ \frac{\partial f}{\partial x}\dfrac{1}{(1-x)^2} \neq \dfrac{0 *(x^2-2x+1) - (2x-2)(1)}{(x^2-2x+1)*(1-x)} $
Why can't I solve the binomial formula in the denominator and then do the derivation?
Thanks in advance!
$\endgroup$ 33 Answers
$\begingroup$There are some small errors in both derivatives. First, I would look ot the power rule first, and will add that.
Power rule
$\frac {d}{dx} (1-x)^{-2} = -2(1-x)^{-3}(-1) = \frac {2}{(1-x)^3}$
Quotient rule
$\frac {d}{dx} \frac {1}{(1-x)^{2}} = \frac {(1-x)^2(\frac {d}{dx} 1) - (1)(\frac {d}{dx}(1-x)^2)}{(1-x)^4}= \frac {- (1)(2)(1-x)(-1)}{(1-x)^4} = \frac {2}{(1-x)^3}$
Multiply the binomial and apply the quotient rule.
$\frac {d}{dx} \frac {1}{1-2x + x^2} = \frac {(1-2x+ x^2)*0 - (-2+2x)}{(1-2x + x^2)^2} = \frac {2(1-x)}{(1-x)^4} = \frac {2}{(1-x)^3}$
$\endgroup$ $\begingroup$By the Quotient Rule, $f'(x)=(-2x+2)\div(1-x)^4$.
It looks like you can expand out $(1-x)^2$, but if you do, the denominator should be $(x^2-2x+1)^2$, which is the same thing as $(1-x)^4$.
Also, I don't think that the partial derivative notation should be used here since $f$ is a function of one variable.
$\endgroup$ $\begingroup$Why you simply use the chain rule?
This formula is basic: for a differentiable function $u(x)$,
$$\biggl(\frac 1{u^n(x)}\biggr)'=\frac{-n}{u^{n+1}(x)}\,u'(x).$$
$\endgroup$
