Find f '(c) by forming the difference quotient [f(c+h)-f(c)]/h and taking the limit as h=>0.

f(x)=2x^3 +1 ; c=1

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To calculate the value of the first derivative in a given point, c = 1, we'll have to apply the limit of the ratio:

limit [f(x) - f(1)]/(x-1), when x tends to 1.

We'll substitute f(x) and we'll calculate the value of f(1):

f(1) = 2*1^3 + 1

f(1) = 2+1

f(1) = 3

limit [f(x) - f(1)]/(x-1) = lim (2*x^3 + 1 - 3)/(x - 1)

We'll combine like terms:

lim (2*x^3 + 1 - 3)/(x - 1) = lim (2*x^3 - 2)/(x - 1)

We'll factorize the numerator by 2:

lim (2*x^3 - 2)/(x - 1) = lim 2(x^3-1)/(x-1)

We'll write the difference of cubes as a product:

x^3 - 1 = (x-1)(x^2 + x + 1)

lim 2(x^3-1)/(x-1) = 2 lim (x-1)(x^2 + x + 1)/(x-1)

We'll simplify the ratio and we'll get:

2 lim (x-1)(x^2 + x + 1)/(x-1) = 2 lim (x^2 + x + 1)

We'll substitute x by 1 and we'll get:

2 lim (x^2 + x + 1) = 2(1^2 + 1 + 1)

2 lim (x^2 + x + 1) = 2*3

2 lim (x^2 + x + 1) = 6

But f'(c) = f'(1) = limit [f(x) - f(1)]/(x-1)

**f'(1) = 6**

To find f'(c) at c= 1 for the function f(x) = 2x^3+1.

Procedure:

f'(x) = lt h-->0 {f(x+h)-f(x)}/h

f'(x) = lt {

f'(x) = lt{(2(x+h)^3+1)- (2x^3+1)}/H

f'(X) = lt{2(x^3+3x^2xh+3xh^2+h^3+1)-(2x^3+1)}/h

f'(x) =lt {3x^2h+3x^2h^2+h^3}/h , as other terms gets cancelled.

f'(x) = Lt3x+ h(3x^2h+h^2) = 3x +0*(3x*0+0^2)

f'(x) =3x.

f'(c) = 3c

f'(1) = 3*1 = 3.

f'(x) = 3xh/h

If f(x) = 2x^3 +1

f'(x) = lim(h-->0) [f(x+h) - f(x)]/ h

=> lim(h-->0)[ 2*(x+h)^3 +1 - 2x^3 -1]/h

=> lim(h-->0)[ 2*(x^3 + 3x^2h+3xh^2+h^3) +1 - 2x^3 -1]/h

=>lim(h-->0)[ 2x^3 + 6x^2h+6xh^2+2h^3 +1 - 2x^3 -1]/h

=>lim(h-->0)[ 6x^2h+6xh^2+2h^3]/h

=>lim(h-->0)[ 6x^2+6xh+2h^2]

=> 6x^2

f'(c) for c=1

=> 6c^2 = 6*1= 6

**Therefore ****for c= 1, ****f'(c)=6.**

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