The definition of derivative of a function $f(x)$ is $$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$

Using this definition, the derivative of $\sin\sqrt{x}$ will be:

$$\lim_{h\to0} \frac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}$$

$$\lim_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$

Now i got stuck. How to find the limit or simplify this expression?

I get intuition that we have to use $$\lim_{x\to0}\frac{\sin x}{x} = 1$$

but that too is leading no where. I am unable to remove h from denominator.


I know the derivative of $\sin\sqrt{x}$ is $\frac{\cos\sqrt{x}}{2\sqrt{x}}$ using chain rule, but this exercise was given to us for practice using division quotient.


$$\lim_{h\to 0}\frac{\sin\sqrt{x+h}-\sin\sqrt x}{h}=\lim_{h\to 0}\frac{\sin\sqrt{x+h}-\sin\sqrt x}{\sqrt{x+h}-\sqrt x}\cdot \frac{\sqrt{x+h}-\sqrt x}{h}$$ $$\underset{u=\sqrt{x+h}}{=}\underbrace{\lim_{u\to \sqrt x}\frac{\sin(u)-\sin(\sqrt x)}{u-\sqrt x}}_{=\cos(\sqrt x)}\cdot \underbrace{\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt x}{h}}_{=\frac{1}{2\sqrt x}}=\frac{\cos(\sqrt x)}{2\sqrt x}.$$

  • $\begingroup$ thank you! can you explain how we get $cos\sqrt{x}$ in leftmost limit of line 2? $\endgroup$
    – integer
    Jan 27 '16 at 14:24
  • $\begingroup$ Using the method that you used in your original post, i.e. use the fact that $$\sin(a)-\sin(b)=2\cos(\frac{a+b}{2})\sin(\frac{a-b}{2}).$$ $\endgroup$
    – Surb
    Jan 27 '16 at 14:26

You got stuck here

$$2 \cdot \lim_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$

You can break that one limit up into the product of two limits.

$$\lim_{h\to 0} \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right) \cdot\lim_{h\to 0} \frac{\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$

The first limit is easy.

$$\lim_{h\to 0} \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right) = \cos\left(\frac{\sqrt{x}+\sqrt{x}}{2}\right) = \cos \sqrt x$$

The second isn't that much harder.

\begin{align} \lim_{h\to 0} \frac{\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h} &= \lim_{h\to 0} \frac{\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)} {\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)} \cdot \frac{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h} \\ &= \lim_{h\to 0} \left(1 \cdot \dfrac{\sqrt{x+h}-\sqrt{x}}{2h}\right) \\ &= \lim_{h\to 0} \dfrac{(x+h)-x}{2h(\sqrt{x+h}+\sqrt{x})} \\ &= \dfrac{1}{4\sqrt{x}} \\ \end{align}

Putting it all together, you get $\dfrac{\cos\sqrt{x}}{2\sqrt{x}}$


You are right. First multiply and divide by the expression in $\sin()$ $$\lim_{h\rightarrow 0} \dfrac{2\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}\times\dfrac{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}$$

Now, as $h \rightarrow 0, \ \ \ \left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)\rightarrow 0$

So we can use $\ \ \lim_{\lambda \rightarrow 0} \dfrac{\sin\lambda}{\lambda} = 1$

So it reduces to: $$\lim_{h\rightarrow 0} \dfrac{\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)}{h}\times\left(\sqrt{x+h}-\sqrt{x}\right)$$

Multiply and divide by $\left(\sqrt{x+h}+\sqrt{x}\right)$

$$\lim_{h\rightarrow 0} \dfrac{\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)}{h}\times\dfrac{\left((\sqrt{x+h})^2-(\sqrt{x})^2\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)}$$

$$ = \lim_{h\rightarrow 0} \dfrac{\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)}{h}\times\dfrac{h}{\left(\sqrt{x+h}+\sqrt{x}\right)}$$

Apply the limit and you get :



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