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Find the equation for the line tangent to the function $f(x) = 2{\sqrt x}$ at $x=25$ Find the slope in two ways:

(i) by using the limit definition of the derivative,

(ii) and using derivative shortcut formulas.

I think was able to find that $p_1= (25,10)$ and bring to use derivative definition, but when I multiply by a "disguised $1$" ,$\frac{2{\sqrt {25+h}}-10}{2{\sqrt {25+h}}-10}$ I'm lost

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  • $\begingroup$ calculate this limit $$ \lim_{h \to 0} \frac{2 \sqrt{x+h}- 2\sqrt x}{h}$$ then put $x=25$. That gives you slope. $\endgroup$ – Santosh Linkha Feb 8 '18 at 4:53
  • $\begingroup$ mathjax references to help you typeset maths. $\endgroup$ – Siong Thye Goh Feb 8 '18 at 4:54
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\begin{align}\lim_{h \to 0}\frac{2\sqrt{25+h} - 10}{h} &= \lim_{h \to 0}\frac{2\sqrt{25+h} - 10}{h} \cdot \frac{2\sqrt{25+h}+10}{2\sqrt{25+h}+10} \\ &=\lim_{h \to 0} \frac{(2\sqrt{25+h})^2-10^2}{h(\sqrt{25+h}+10)}\\ &=\lim_{h \to 0} \frac{4(25+h)-100}{h(\sqrt{25+h}+10)}\\\end{align}

Can you finish this?

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