Question

Use the first principle to differentiate? `y=sqrt(sinx)`

Answer 1

Step one is to rewrite the function as a rational exponent `f(x) = sin(x)^{1/2}`

Explanation:

After you have your expression in that form, you can differentiate it using the Chain Rule:

In your case: `u^{1/2} -> 1/2Sin(x)^{-1/2}*d/dxSin(x)`

Then, `1/2Sin(x)^{-1/2}*Cos(x)` which is your answer

Answer 2

`d/dx sqrt(sinx) = cosx/(2sqrt(sinx))`

Explanation:

Using the limit definition of the derivative we have:

`f'(x) = lim_(h rarr 0) (f(x+h)-f(x)) / (h)`

So for the given function, where `f(x)=sqrt(sinx)`, we have:

`f'(x) = lim_(h rarr 0) (sqrt(sin(x+h))-sqrt(sinx)) / (h)`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) (sqrt(sin(x+h))-sqrt(sinx)) / (h) * (sqrt(sin(x+h))+sqrt(sinx))/(sqrt(sin(x+h))+sqrt(sinx))`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) (sin(x+h)-sinx) / (h(sqrt(sin(x+h))+sqrt(sinx)))`

Then we can use the trigonometric identity:

`sin(A+B) -= sinAcosB + cosAsinB`

Giving us:

`f'(x) = lim_(h rarr 0) (sinxcos h+cosxsin h-sinx) / (h(sqrt(sin(x+h))+sqrt(sinx)))`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) (sinx(cos h-1)+cosxsin h) / (h(sqrt(sin(x+h))+sqrt(sinx)))`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) (sinx(cos h-1)) / (h(sqrt(sin(x+h))+sqrt(sinx))) + (cosxsin h) / (h(sqrt(sin(x+h))+sqrt(sinx)))`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) (cos h-1)/h (sinx) / (sqrt(sin(x+h))+sqrt(sinx)) + (sin h)/h (cosx) / (sqrt(sin(x+h))+sqrt(sinx))`

Then we use two very standard calculus limits:

`lim_(theta -> 0) sintheta/theta =1`, and `lim_(theta -> 0) (costheta-1)/theta =0`, and #

And we can now evaluate the limits:

`f'(x) = 0 xx (sinx) / (sqrt(sin(x))+sqrt(sinx)) + 1 xx (cosx) / (sqrt(sin(x))+sqrt(sinx))`

`\ \ \ \ \ \ \ \ \ = (cosx) / (2sqrt(sin(x))`