Question

Diffentiate with respect to $x$:-
i) $\frac{x+1}{x+2}\left[\frac{1}{{\left(x+2\right)}^2}\right]$
ii) $x{e}^x\left[\left(x+1\right)e^x\right]$

Answer 1

Part (1)

$y=\frac{x+1}{x+2}\left(\frac{1}{{(x+2)}^2}\right)$

$y=\frac{x+1}{{(x+2)}^3}$

On differentiating both sides w.r.t $x$, we get

$\frac{dy}{dx}=\frac{d}{dx}\left[\frac{x+1}{{(x+2)}^3}\right]$

$\frac{dy}{dx}=\left[\frac{{(x+2)}^3(1+0)-(x+1)\left(3{(x+2)}^2(1+0)\right)}{{\left[{(x+2)}^3\right]}^2}\right]$

$\frac{dy}{dx}=\frac{{(x+2)}^3-3(x+1){(x+2)}^2}{{(x+2)}^6}$

$\frac{dy}{dx}=\frac{(x+2)-3(x+1)}{{(x+2)}^4}$

$\frac{dy}{dx}=\frac{x+2-3x-3}{{(x+2)}^4}$

$\frac{dy}{dx}=\frac{-2x-1}{{(x+2)}^4}$

$\frac{dy}{dx}=\frac{-(2x+1)}{{(x+2)}^4}$

Hence, this is the answer.

Part (2)

$y=x{e}^x\left[(x+1)e^x\right]$

$y=x{e}^{2x}(x+1)$

$y=x^2{e}^{2x}+x{e}^{2x}$

On differentiating both sides w.r.t $x$, we get

$\frac{dy}{dx}=\frac{d}{dx}(x^2{e}^{2x}+x{e}^{2x})$

$\frac{dy}{dx}=x^2{e}^{2x}\left(2\right)+e^{2x}\left(2x\right)+x{e}^{2x}\left(2\right)+e^{2x}\left(1\right)$

$\frac{dy}{dx}=2x^2{e}^{2x}+2x{e}^{2x}+2x{e}^{2x}+e^{2x}$

$\frac{dy}{dx}=2x^2{e}^{2x}+4x{e}^{2x}+e^{2x}$

$\frac{dy}{dx}=e^{2x}(2x^2+4x+1)$

Hence, this is the answer.