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

2 Answers
Mar 18, 2018

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

Mar 18, 2018

# 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)) #