Hello
can someone please show me how to do this sum correct
I need to find the derivative of x^-4(1-X^-4)^3
the correct answer is in yellow at the bottom,my answer is above it
please see attached photo thanks
$\displaystyle x^{-4}(1-x^{-4})^3 = \frac{(x^4-1)^3}{x^{16}}$
$\displaystyle \frac{d}{dx}\left[\frac{(x^4-1)^3}{x^{16}}\right]$
$\displaystyle \frac{x^{16} \cdot 3(x^4-1)^2 \cdot 4x^3 - (x^4-1)^3 \cdot 16x^{15}}{x^{32}}$
$\displaystyle \frac{4x^{15}(x^4-1)^2 [3x^4 - 4(x^4-1)]}{x^{32}}$
$\displaystyle \frac{4x^{15}(x^4-1)^2 [4-x^4]}{x^{32}}$
$\displaystyle \frac{4(x^4-1)^2 (4-x^4)}{x^{17}}$
Looks like an application of the product and chain rule
$\displaystyle x^{-4}(1-x^{-4})^3$
$\displaystyle u = x^{-4} \: \: \rightarrow \: \: u' = -4x^{-3}$
$\displaystyle v = (1-x^{-4})^3 \: \: \rightarrow \:\: v' = 3(1-x^{-4})^2 \cdot 4x^{-3} = 12x^{-3}(1-x^{-4})^2$
$\displaystyle y' = -4x^{-3}(1-x^{-4})^3 + 12x^{-7} \cdot (1-x^{-4})^2$
$\displaystyle y' = -4x^{-3}[(1-x^{-4})^3 + 3x^{-4}(1-x^{-4})^2]$
Why they want you to use the binomial expansion is beyond me but it's as follows up to $\displaystyle x^3$
$\displaystyle (1+x)^n = 1+nx+\frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3$