Question

d/dx(e^x.x^7)=?plzz give solution fast​

Answer 1

Let the given function be f(x) .

$\therefore \: \: y = f(x)$

$= > y = {e}^{x} \times {x}^{7}$

$= > \dfrac{dy}{dx} = \dfrac{d \bigg \{ {e}^{x} \times {x}^{7} \bigg \} }{dx}$

Applying Product Rule Of Differentiation :

$= > \dfrac{dy}{dx} = {x}^{7} \times \dfrac{d \{ {e}^{x} \}}{dx} + {e}^{x} \times \dfrac{d \{ {x}^{7} \}}{dx}$

$= > \dfrac{dy}{dx} = {x}^{7} \times {e}^{x} + {e}^{x} \times (6 {x}^{6} )$

Taking common terms :

$= > \dfrac{dy}{dx} = {e}^{x} {x}^{6} \bigg \{x + 6 \bigg \}$

So final answer is :

$\boxed{ \green{ \bold{\huge{ \dfrac{dy}{dx} = {e}^{x} {x}^{6} \bigg \{x + 6 \bigg \}}}}}$

Basic rules for Differentiation applied here :

$1) \: \: f(x) = u \times v$

$= > \dfrac{d \{f(x) \}}{dx} = u \dfrac{dv}{dx} + v \dfrac{du}{dx}$

$2) \: \: f(x) = {x}^{n}$

$= > \dfrac{d \{f(x) \}}{dx} = n {x}^{n - 1}$

Answer 2

Let ,

$\sf y = {(e)}^{x} \times {(x)}^{7}$

Differentiating with respect to x , we get

$\sf \mapsto \frac{d(y)}{dx} = \frac{d({(e)}^{x} \times {(x)}^{7} )}{dx} \\ \\ \sf \mapsto \frac{d(y)}{dx} = {(e)}^{x} \times \frac{d {(x)}^{7} }{dx} + {(x)}^{7} \times \frac{d {(e)}^{x} }{dx} \: \: \: \: \{ \because product \: rule\} \\ \\ \sf \mapsto \frac{d(y)}{dx} = {(e)}^{x} \times 7 {(x)}^{6} + {(x)}^{7} \times {(e)}^{x} \: \: \: \{ \because \frac{d {(e)}^{x} }{x} = {(e)}^{x} \}\\ \\ \sf \mapsto \frac{d(y)}{dx} =7 {(x)}^{6} {(e)}^{x} + {(x)}^{7} {(e)}^{x} \\ \\ \sf \mapsto \frac{d(y)}{dx} = {(e)}^{x} {(x)}^{6} \bigg (7 + x\bigg)$

$\sf \therefore{ \underline{the \: value \: of \: \frac{d(y)}{dx} \: is \: {(e)}^{x} {(x)}^{6} \bigg (7 + x\bigg)}}$