Question

From first principle, find the derivative of y=2x^3+x²+4x with respect to X​

Answer 1

`derivative \: of \: y = 6 {x}^(2) + 2x + 4`

Explanation:

we know that

`(d)/(dx) {x}^(n) = n {x}^(n - 1)`

Now the derivative of y=2x³+x²+4x

`(dy)/(dx) = (d)/(dx) (2 {x}^(3) + {x}^(2) + 4x)`

`(dy)/(dx) = (d)/(dx) (2 {x}^(3) ) + (d)/(dx)( {x}^(2) ) + (d)/(dx) (4x)`

`(dy)/(dx) = 2((d)/(dx) {x}^(3) ) + (d)/(dx) {x}^(2) + 4((d)/(dx) x)`

here

`(d)/(dx) {x}^(3) = 3 {x}^(2)`

`(d)/(dx) {x}^(2) = 2x`

`(d)/(dx) x = 1`

Now

`(dy)/(dx) = 2(3 {x}^(2) ) + 2x + 4(1)`

`(dy)/(dx) = 6 {x}^(2) + 2x + 4`

I hope it helped you