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

[kashan123999]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

 

[arildno]

Rewrite:
\sin(\sqrt{x+Dx})=\sin(\frac{\sqrt{x}}{\sqrt{1+\frac{Dx}{\sqrt{x}}}}<br /> +\frac{\frac{Dx}{\sqrt{x}}}{\sqrt{1+\frac{Dx}{\sqrt{x}}}})

 

[kashan123999]

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

 

[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.