f(x) = x^2 -3 ln x fin f'(1)

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We have f(x) = x^2 - 3ln x.

The derivative of f(x) can be found by using the relations that the derivative of x^n is nx^(n-1) and derivative of ln x is 1/x

f'(x) = 2x - 3/x

f'(1) = 2*1 - 3/1

= 2- 3

= -1

**The required value of f' (1) is -1**

We'll calculate the expression of f'(x) first:

f'(x) = (x^2 -3 ln x)'

f'(x) = 2x - 3/x

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

f'(1) = 2*1 - 3/1

f'(1) = 2 - 3

**f'(1) = -1**

**Another method is to calculate the limit of the ratio:**

**lim [f(x) - f(1)/(x-1)],when x->1**

f(x) = x^2 -3 ln x fin f'(1)

f'(x) = {x^2-3lnx}'

= (x^2)' - (3lnx)'

f'(x)= 2x-3/x

f'(1) = 2*1 -3/1 = 2-3

f'(1) = 2- 3 =1

f(x) = x^2 - 3lnx

First we will determine f'(x).

f'(x) = (x^2)' - 3(lnx)'

= 2x - 3*(1/x)

= 2x - 3/x

Now we will substitute x= 1

f'(1) = 2 - 3 = -1

**==> f'(1) = -1**

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