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

Question

Ezra Adams · Accepted Answer

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

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

Samantha Adkison · Answer

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

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)) #

Samantha Adkison · Answer

undefined To differentiate the function y = √(sinx) using the first principle:

Let's denote the given function as y = u^v, where u = sinx and v = 1/2.

Now, we'll use the first principle of differentiation:

dy/dx = lim(h->0) [(√(sin(x + h)) - √(sinx)) / h]

= lim(h->0) [(√(sin(x + h)) - √(sinx)) / h] * [(√(sin(x + h)) + √(sinx)) / (√(sin(x + h)) + √(sinx))]

= lim(h->0) [(sin(x + h) - sinx) / (h * (√(sin(x + h)) + √(sinx)))]

= lim(h->0) [(sinx*cos(h) - sinx) / (h * (√(sinx*cos(h) + sin^2(x)))]

= lim(h->0) [(sinx * (cos(h) - 1)) / (h * (√(sinx*cos(h) + sin^2(x))))]

= sinx * lim(h->0) [(cos(h) - 1) / h] / (√(sinx))

Now, we know that lim(h->0) [(cos(h) - 1) / h] = 0.

Therefore, dy/dx = 0 / (√(sinx))

= 0

So, the derivative of y = √(sinx) with respect to x using the first principle is 0.