Question

please help my guys for differentiate with respect to x​

Answer 1

Step-by-step explanation:

(a)

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

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

$+ 3. 2.{x}^{2 - 1} - 2 = > 4 {x}^{3} + 6x - 2$

(b)

$\frac{d}{dx}( {x}^{2}cosx) = > cosx \frac{d}{dx}( {x}^{2}) +$

${x}^{2} \frac{d}{dx}(cosx) = > 2xcosx - {x}^{2}sinx$

(c)

let's take

${(6x + 7)} \: as \: y \:$

So

$\frac{d}{dx} {(y}^{4}) = > 4 {y}^3 \frac{dy}{dx} = > 4. {(6x + 7)}^{3}.6$

$= > 24 {(6x + 7)}^{3}$

(d)

$\frac{d}{dx}( {x}^{5} {e}^{x}) = > {e}^{x} \frac{d}{dx}( {x}^{5}) + {x}^{5} \frac{d}{dx}( {e}^{x}) = >$

$5 {x}^{4} {e}^{x} + {x}^{5} {e}^{x} = > {e}^{x}(5 {x}^{4} + {x}^5)$

(e)

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

$= > - \frac{x}{ {e}^{x} }$

Hope it helps you.