Fiding a derivative

Question

I need to find the derivative of $\sqrt{x^2+3x}$ using the definition of derivative. e.g. $\frac{f(x)-f(a)}{x-a}$ as x->a. Normally I get these but the $x^2$ is messing me up. I am at $$\lim \limits_{x \to a}\frac{3(x-a) + x^2 - a^2}{(x-a)(\sqrt{x^2+3x}+\sqrt{a^2+3a})}$$ But I don't know how I should clear the $+x^2 - a^2$

Answer 1

Continuing from the point you're at, note that $x^2-a^2 = (x+a)(x-a)$.

So, we have: $$\lim \limits_{x \to a}\frac{3(x-a) +(x+a)(x-a)}{(x-a)(\sqrt{x^2+3x}+\sqrt{a^2+3a})}$$

Cancel out the $(x-a)$ terms: $$\lim \limits_{x \to a}\frac{3 +(x+a)}{(\sqrt{x^2+3x}+\sqrt{a^2+3a})}$$

Now, just evaluate the limit: $$\lim \limits_{x \to a}\frac{3 +(x+a)}{(\sqrt{x^2+3x}+\sqrt{a^2+3a})}$$ $$=\frac{3 +(x+x)}{(\sqrt{x^2+3x}+\sqrt{x^2+3x})}$$ $$=\boxed{\dfrac{2x+3}{2\sqrt{x^2+3x}}}$$