# `g(x) = 2x^2 - 3x, c=2` Use the alternative form of the derivative to find the derivative at x = c (if it exists). Given the function `f(x)=2x^2-3x` , c=2.

We have to use the alternative form of derivative to find the derivative at x=c.

Here given that c=2.

So by definition of alternative derivative we have,

`f'(c)=lim_(x->c)(f(x)-f(c))/(x-c)`

`f'(2)=lim_(x->2)(f(x)-f(2))/(x-2)`

` =lim_(x->2)(2x^2-3x-(2(2^2)-3(2)))/(x-2)`

`=lim_(x->2)(2x^2-3x-2)/(x-2)`

`=lim_(x->2)((2x+1)(x-2))/(x-2)`

`=lim_(x->2) 2x+1`

`=5`

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Given the function `f(x)=2x^2-3x` , c=2.

We have to use the alternative form of derivative to find the derivative at x=c.

Here given that c=2.

So by definition of alternative derivative we have,

`f'(c)=lim_(x->c)(f(x)-f(c))/(x-c)`

`f'(2)=lim_(x->2)(f(x)-f(2))/(x-2)`

` =lim_(x->2)(2x^2-3x-(2(2^2)-3(2)))/(x-2)`

`=lim_(x->2)(2x^2-3x-2)/(x-2)`

`=lim_(x->2)((2x+1)(x-2))/(x-2)`

`=lim_(x->2) 2x+1`

`=5`

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