Question

Find the derivative with the formula: f’(a)=lim x-a f(x)-f(a)/x-aQuestion: f(x)=3x^2+x

Answer 1

Ok, so

Given the alternate definition of deritative:

$f^{\prime}(a)=\lim _{x\to a}(\frac{f(x)-f(a)}{x-a})$

We want to find the deritative of the function:

$f(x)=3x^2+x$

For this, we could start by replacing:

$f^{\prime}(a)=\lim _{x\to a}(\frac{3x^2+x-(3a^2+a)}{x-a})$

Now, we could distribute the sign outside the bracket:

$f^{\prime}(a)=\lim _{x\to a}(\frac{3x^2+x-(3a^2+a)}{x-a})=\lim _{x\to a}(\frac{3x^2+x-3a^2-a)}{x-a})$

And, as you can see, this expression can be even more simplified if we factor. We have to ask for a way to eliminate the indeterminacy caused by the denominator:

$f^{\prime}(a)=\lim _{x\to a}(\frac{3(x^2-a^2)+(x-a)}{x-a})$

Now, remember that:

$x^2-a^2=(x+a)(x-a)$

So, our expression will be:

$f^{\prime}(a)=\lim _{x\to a}(\frac{3(x^2-a^2)+(x-a)}{x-a})=\lim _{x\to a}(\frac{3(x-a^{})(x+a)+(x-a)}{x-a})$

We could group some terms:

$f^{\prime}(a)=\lim _{x\to a}(\frac{3(x-a^{})(x+a)+(x-a)}{x-a})=\lim _{x\to a}(\frac{(x-a)\lbrack3(x+a)+1\rbrack}{x-a})$

As you can see, the term (x-a) can be cancelled. And, our limit will be:

$\begin{gathered} f^{\prime}(a)=\lim _{x\to a}(3(x+a)+1)=(3(a+a)+1) \\ f^{\prime}(a)=3(2a)+1 \\ f^{\prime}(a)=6a+1 \end{gathered}$

Therefore, f'(a)=6a+1