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

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10 votes

Answer:


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

User Petrusqui
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