Differentiate

Solution:

To find: Differentiation of $\frac{e^{x}(x-1)}{(x+1)}$

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d e^{x}}{d x}=e^{x}$

(iii) $\frac{d x^{n}}{d x}=n x^{n-1}$

(iv) (uv)′ = u′v + uv′ (Leibnitz or product rule)

Let us take $u=e^{x}(x-1)$ and $v=(x+1)$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left[e^{x}(x-1)\right]}{d x}$

Applying Product rule

$(g h)^{\prime}=g^{\prime} h+g h^{\prime}$

Taking $g=e^{x}$ and $h=x-1$

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

$=e^{x}(x-1)+e^{x}$

$u^{\prime}=e^{x} x$

$v^{\prime}=\frac{d v}{d x}=\frac{d(x+1)}{d x}=1$

Putting the above obtained values in the formula:-

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

$\left[\frac{e^{x}(x-1)}{(x+1)}\right]^{\prime}=\frac{\left(e^{x} x\right)(x+1)-\left[e^{x}(x-1)\right](1)}{(x+1)^{2}}$

$=\frac{e^{x} x^{2}+e^{x} x-e^{x} x+e^{x}}{(x+1)^{2}}$

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

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

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