Derivative using limit definition

[AvgStudent]

Find [imath]\boxed{\frac{d}{dx}(\sin \sqrt{x})}[/imath] using the definition of derivative.
Hint: Use the following facts
[math]\boxed{1)\,\sin(A)-\sin(B)=2\sin\Bigg(\frac{A-B}{2}\Bigg)⋅\cos\Bigg(\frac{A+B}{2}\Bigg)}\\ \boxed{2)\, \lim_{\theta \to 0}\,\frac{\sin\theta}{\theta}=1}[/math]What I have so far...
[math]\lim_{h \to 0} \frac{\sin(\sqrt{x}+h)- \sin{\sqrt{x}}}{h}= 2\cdot\lim_{h \to 0} \frac{\sin(\frac{\sqrt{x+h}-\sqrt{x}}{2})\cos(\frac{\sqrt{x+h}+\sqrt{x}}{2})}{h}[/math]I'm supposed to use the second fact here somehow but don't see it. Help, please

 

[blamocur]

First hint:
[math]\lim_{h\rightarrow 0} \frac {\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)} {h} = \lim_{h\rightarrow 0} \frac {\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)} {\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)} \cdot \frac {\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)} {h}[/math]

 

[AvgStudent]

[math]2\lim_{h\rightarrow 0} \frac {\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)} {\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)} \cdot \frac {\left(\sqrt{x+h}-\sqrt{x}\right)} {2h} \cdot \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)= (1)\cos\sqrt{x}\lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}[/math]How to do the last limit? I don't want to use l'Hospital rule and do derivative

 

[blamocur]

[math]2\lim_{h\rightarrow 0} \frac {\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)} {\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)} \cdot \frac {\left(\sqrt{x+h}-\sqrt{x}\right)} {2h} \cdot \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)= (1)\cos\sqrt{x}\lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}[/math]How to do the last limit? I don't want to use l'Hospital rule and do derivative

Multiply both the enumerator and the denominator by [imath]\sqrt{x+h}+\sqrt{x}[/imath]

 

[Steven G]

You can't use l'hopital's rule since you were required to use the definition of derivative.

Just multiply numerator and denominator by the conjugate of the numerator.

 

[AvgStudent]

You can't use l'hopital's rule since you were required to use the definition of derivative.

Just multiply numerator and denominator by the conjugate of the numerator.

Right...?

Multiply both the enumerator and the denominator by [imath]\sqrt{x+h}+\sqrt{x}[/imath]

Yes, good trick.
[math]\cos\sqrt{x}\lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}} {\sqrt{x+h}+\sqrt{x}} = \cos\sqrt{x} \lim_{h\to 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})} = \frac{\cos\sqrt{x}}{2\sqrt{x}}\quad \blacksquare[/math]Thanks, both