Question

Find the derivative of sin√x

Answer 1

Hey,

d/dX sin√x

cos√x.d/dx √x

cos√x. 1/2 x^-1/2-1

cos√x/2√x.

HOPE IT HELPS YOU:-))

Answer 2

Hey there!

Solution:


Let y = sin$\sqrt{x}$


Put $\sqrt{x}$ = t

y = sin t


Now differentiate both sides wrt x;


dy/dx = (cos t) . (dt/dx).... ( 1 )


Now finding value of dt/dx:


t = $\sqrt{x}$


t = $x^{1/2}$


Now again differentiate both sides wrt x:

dt/dx = $\dfrac{1}{2\sqrt{x}}$


use this value in equation ( 1 ).


dy/dx = cos t × $\dfrac{1}{2\sqrt{x}}$


Substituting value of t and we get;


dy/dx = $\dfrac{1}{2\sqrt{x}}$ × cos$\sqrt{x}$


#Be Brainly.