# How to calculate Derivative of sin sq. root x by definition?

1. Sep 19, 2013

### kashan123999

1. The problem statement, all variables and given/known data

Evaluate derivative of (sin sq. root x) w.r.t x?

2. Relevant equations

Limit Δx--> 0 (sin√(x+Δx) - sin(√x)) / Δx

3. The attempt at a solution

i couldn't operate it from here.... Δy = (2cos((√x+Δx) + (√x)) . sin((√x+Δx) - (√x)) / Δx...???

2. Sep 19, 2013

### arildno

Rewrite:
$$\sin(\sqrt{x+Dx})=\sin(\frac{\sqrt{x}}{\sqrt{1+\frac{Dx}{\sqrt{x}}}} +\frac{\frac{Dx}{\sqrt{x}}}{\sqrt{1+\frac{Dx}{\sqrt{x}}}})$$

Last edited: Sep 19, 2013
3. Sep 19, 2013

### kashan123999

where did that sq. root (1+ Dx/sq.root x) come from?

4. Sep 19, 2013

### arildno

Multiply the argument within the sine as follows:
$$\sqrt{x+Dx}=\sqrt{x+Dx}*1=(\sqrt{x+Dx})*\frac{\sqrt{x+Dx}}{\sqrt{x+Dx}}=\frac{x+Dx}{\sqrt{x+Dx}}$$
Now, extraxt "x" from the square root in the denominator and simplify.