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

  • #1

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...???
 

Answers and Replies

  • #2
arildno
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Rewrite:
[tex]\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}}}})[/tex]
 
Last edited:
  • #3
where did that sq. root (1+ Dx/sq.root x) come from?
 
  • #4
arildno
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Multiply the argument within the sine as follows:
[tex]\sqrt{x+Dx}=\sqrt{x+Dx}*1=(\sqrt{x+Dx})*\frac{\sqrt{x+Dx}}{\sqrt{x+Dx}}=\frac{x+Dx}{\sqrt{x+Dx}}[/tex]
Now, extraxt "x" from the square root in the denominator and simplify.
 

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