[tex]\frac{dy}{dx}[/tex] = 2a/b²(x + 1)
y = a/u² = a/ (b²/(x + 1)²) = a/b² (x + 1)²
differentiate using the ' chain rule'
[tex]\frac{dy}{dx}[/tex] = 2a/b² (x + 1)
First of all, substitute [tex] u^2 [/tex] in the first expression:
[tex] u = \dfrac{b}{x+1} \implies u^2 = \dfrac{b^2}{(x+1)^2} [/tex]
So, your function is
[tex] y = \dfrac{a}{\frac{b^2}{(x+1)^2}} = \dfrac{a(x+1)^2}{b^2} = \dfrac{a}{b^2}(x+1)^2 [/tex]
Since a and b are constants, we have
[tex] \dfrac{dy}{dx} \dfrac{a}{b^2}(x+1)^2 = \dfrac{a}{b^2}\dfrac{dy}{dx}(x+1)^2 [/tex]
The derivative of [tex] (x+1)^2 [/tex] is
[tex] 2(x+1)\cdot 1 [/tex]
following the rule
[tex] \dfrac{d}{dx} f^n(x) = nf^{n-1}(x)f'(x) [/tex]
So, the answer is
[tex] \dfrac{dy}{dx} \dfrac{a}{b^2}(x+1)^2 = \dfrac{2a}{b^2}(x+1) [/tex]
