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1 vote
1 vote
Differentiate

y = 3x {}^(3) + 8x - 7





User Syldor
by
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2 Answers

12 votes
12 votes

Answer:

9x² + 8

Explanation:

The given function to us is ,


\implies y = 3x {}^(3) + 8x - 7

And we need to differentiate the given function with respect to x . Taking the given function and differenciating wrt x , we have


\implies y = 3x^3 + 8x - 7

Recall that , the derivative of constant is 0 . Therefore ,


\implies (dy )/(dx)= (d)/(dx)(3x^3 + 8x - 7) \\\\\implies(dy )/(dx)= (d)/(dx)(3x^3)+(d)/(dx)(8x) + 0 \\\\\implies(dy )/(dx)= 3* 3 . x^(3-1) + 8* 1 . x^(1-1) \\\\\implies\underline{\underline{(dy )/(dx)= 9x^2+8}}

Hence the derivative of given function is 9x² + 8 .

User Sean Rucker
by
3.1k points
14 votes
14 votes

Answer:

y=9x^2 + 8

Explanation:

using the power rule, we will differentiate each term separately

d/dx of 3x^3 = (3)(3)x^(3-1) = 9x^2

d/dx of 8x = 8x^(1-1) = 8

d/dx of -7 = 0

combining them we get the derivative which is y = 9x^2 + 8

User ItzBenteThePig
by
3.2k points