Question

ANswer Fast...Perfect Ans... ​

Answer 1

Step-by-step explanation:

$\sf \star \: Question \: 7 \: \star \\ \\ \sf y {({x}^{3}-2x-1)}^{5} \\ \\ \sf differentiating \: with \: respect \: to \: x \\ \\ \sf \frac{dy}{dx} = 5{({x}^{3}-2x-1)}^{5-1} × \frac{d({x}^{3}-2x-1)}{dx} \\ \\ \sf \frac{dy}{dx} = 5{({x}^{3}-2x-1)}^{4} × ( 3{x}^{2}-2) \\ \\ \sf \frac{dy}{dx} = 5(3{x}^{2}-2){({x}^{3}-2x-1)}^{4}$

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \star \: your \: answer \: \star$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y{(x}^{3} - 2x - {1}^{5)}$

. differentiating with respect to x

$\frac{dy}{dx} = 5( {x}^{2} - 2x - {5 }^{5 - 1} \times \frac{d( {x}^{3 - 2x - 1}) }{dx}$

$\frac{dy}{dx} = 5( {x}^{3} - 2x - {1}^{4}) \times ({3}^{2} - 2)$

Hope this will help you ...