How do I find the derivative of f(x) = 3x² + x - 7?

Answer 1

To find the derivative of this expression, you can take the derivative of each term separately, one at a time.

You can use a special tool called the Power Rule.

Power Rule: d/dx x^n = nx^(n-1)

Now, let's apply this to your question. First, we'll take the derivative of 3x² using the Power Rule:

d/dx 3x² = 2*3x^(2-1) = 6x

Now, let's find the derivative of x. Keep in mind that x can be written as x^1:

d/dx x^1 = 1*x^(1-1) = x^0 = 1

Now, let's find the derivative of -7. Keep in mind that -7 can be written as -7 * x^0, because x^0 = 1:

d/dx -7 * x^0 = 0 * -7^(0-1) = 0 * -7^(-1) = 0

Now, just add the terms together:

f'(x) = 6x + 1 - 0

f'(x) = 6x + 1

Answer 2

For a sum problem, multiply term coefficient by exponent and drop value of exponent by 1.
Bearing in mind 7 can be written as 7x^0.
dy/dx of 3x^2+x-7 = 2(3x)^(2-1) +1(x)^(1-1)-0(7)^-1
=6x^1+x^0-0
=6x+1

Answer 3

do derivative of each term idividually...

u wll get

3X2 x+ 1+0= 6x+1

Answer 4

derivative of a sum is the sum of the derivative

f'(x) = d/dx (3x^2) + d/dx(x) + d/dx (-7)

d/dx (u)^n = n * (u)^(n - 1) * d/dx(u)

f'(x) = 6x + 1

Answer 5

f(x=3x^2+x-7
f'(x= 6x+1. ANS.

Answer 6

2*3*x^(2-1)+x^(1-1)+0=6x+1

Answer 7

f `(x) = 6x + 1

Answer 8

f'(x) = 6x + 1

d(x^n)/dx = nx^(n-1)
d(constant)/dx = 0