Use the limit definition to find the slope of the tangent line to the graph of $f(x)=\sqrt{8x+1}$ at the point $(6,7)$.
I sort of get how to do this but the square root is making it difficult. I know there is way but I don't remember. Can someone show how to solve this problem?
Best Answer
Hint: After you set it up, make use of this algebraic manipulation:
$$\sqrt {u} - \sqrt {v} \;\; = \;\; \frac{\sqrt {u} - \sqrt {v}}{1} \cdot \frac{\sqrt {u} + \sqrt {v}}{\sqrt {u} + \sqrt {v}} \;\; = \;\; \frac{u-v}{\sqrt {u} + \sqrt {v}} $$
In your case, $u$ will be $8(6+h) + 1$ and $v$ will be $8\cdot6 + 1.$