Use the first principle to differentiate? y=sqrt(sinx)
2 Answers
Step one is to rewrite the function as a rational exponent
Explanation:
After you have your expression in that form, you can differentiate it using the Chain Rule:
In your case:
Then,
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) = 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 , andlim_(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))