Answer

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**Hint:**These types of problems are pretty straight forward and are very simple to solve. For solving problems of this kind, we first need to recall the theory of limits and then try the problem. According to the general theory of limit, if we consider two points on a curve, say \[\left( x,f\left( x \right) \right)\] and \[\left( x+h,f\left( x+h \right) \right)\] , where \[h\] is a infinitesimally small quantity, then the derivative or in general terms the slope of the line is given by,

\[{{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}\] . We now put the value of \[f\left( x \right)\] in the given defined equation to get the value of the derivative.

**Complete step by step answer:**

We now start off with the solution to the given problem as,

We rewrite the value of the derivative of the given function \[f\left( x \right)\] as,

\[{{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}\]

In our given problem, the value of \[f\left( x \right)\] is given to be as \[f\left( x \right)=3x-4\] . Now, replacing this value in our above equation, we then rewrite it as,

\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{\left( 3\left( x+h \right)-4 \right)-\left( 3x-4 \right)}{h}\]

Now, removing the brackets and evaluating we get,

\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{3x+3h-4-3x+4}{h}\]

Now cancelling \[3x\] and \[4\] from the above we get,

\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{3h}{h}\]

Now cancelling \[h\] form the numerator and the denominator, we get,

\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}3\]

This results to the value,

\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=3\]

**Thus, the derivative of the given function is \[{{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=3\] .**

**Note:**For these types of problems, we need to remember the theory of limits and the general form about how the derivative of a function is represented using limits. We can also cross check the answer to the given problem, by using normal differentiation that we often use. This form of solving the problem is a much more fundamental method of solving a calculus problem.

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