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# , 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)) #