I'm trying to find the derivative of:
$$
\frac{d}{dx}\left(\tan \left(\sqrt{x}\right)\right)
$$
As per the chain rule I have to find the derivative of $tan()$ and then $(\sqrt{x})$ which at the end is equal to $\frac{\sec ^2\left(\sqrt{x}\right)}{2\sqrt{x}}$
However, in my exercise, I don't have to use the chain rule nor L'Hopital rule.
I'm looking to re-write $\tan(\sqrt{x})$ in a way to solve the exercise. I'm thinking of
$$
\lim _{h\to 0}\left(\frac{\tan\left(\sqrt{x+h}\right)-\tan\left(\sqrt{x}\right)}{h}\right)
$$
but I'm kind of lost on how to solve it
Best Answer
You should first show your work so others can see where you went wrong. Anyway I think I know where is the mistake as I myself have done same mistake in past. So here is my solution
Using first principle of derivative we can write, $$\frac{d}{dx}\tan(\sqrt{x}) = \lim_{h \to 0} \frac{\tan(\sqrt{x+h})-\tan(\sqrt{x})}{h}$$ $$=\lim_{h \to0} \frac{\sin(\sqrt{x+h})\cos(\sqrt{x})-\cos(\sqrt{x+h})\sin(\sqrt{x})}{h\cos(\sqrt{x+h})\cos(\sqrt{x})}$$ $$=\lim_{h \to0} \frac{\sin(\sqrt{x+h}-\sqrt{x})}{h\cos(\sqrt{x+h})\cos(\sqrt{x})}$$ Applying algebra of limits we can write it as $$=\lim_{h \to0} \frac{\sin(\sqrt{x+h}-\sqrt{x})}{h}.\lim_{h \to 0}\frac{1}{\cos(\sqrt{x+h})\cos(\sqrt{x})}$$ The second limit evaluates to $\sec^2(\sqrt{x})$ and possibly at this point you made the mistake by putting first limit equal to 1. But it is not so. We multiply and divide by $(\sqrt{x+h}-\sqrt{x})$ and get $$\frac{d}{dx}\tan(\sqrt{x}) = \sec^2(\sqrt{x})\lim_{h \to0} \frac{\sin(\sqrt{x+h}-\sqrt{x})}{h}.\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}-\sqrt{x}}$$ $$=\sec^2(\sqrt{x})\lim_{h \to0} \frac{\sin(\sqrt{x+h}-\sqrt{x})}{\sqrt{x+h}-\sqrt{x}}.\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$$
Now it is easy to see that the first limit gives 1 and the second one evaluates to $\frac{1}{2\sqrt{x}}$ finally giving $$\frac{d}{dx}\tan(\sqrt{x}) = \frac{\sec^2(\sqrt{x})}{2\sqrt{x}}$$