How do you differentiate #f(x) = sqrt(sin^3(2-x^2) # using the chain rule?

Question

How do you differentiate #f(x) = sqrt(sin^3(2-x^2)     # using the chain rule?

Cameron Alexander · Accepted Answer

#dy/dx=f'(x)#=#-6xsqrtsin(2-x^2)cos(2-x^2).#

Let #y=f(x)=sqrt(sin^3(2-x^2)#

Put #u=sin^3(2-x^2),# so, #y=sqrtu#

Next, #u=sin^3(2-x^2)={sin(2-x^2)}^3=t^3,# say, where #t=sin(2-x^2)#

Finally, take #2-x^2=v#, so, #t=sinv#

Thus, #y# is a fun. of #u, u# is a fun. of #t, t# is a fun. of #v, v# is a fun. of #x.#

Hence, by Chain Rule,

#dy/dx=dy/(du)(du)/(dt)(dt)/(dv)(dv)/dx...........(1)#

#y=sqrtu rArr dy/du=1/(2sqrtu).#

#u=t^3 rArr du/dt=3t^2.#

#t=sinv rArr dt/dv=cosv.#

#v=2-x^2 rArr dv/dx=-4x.#

Subing all these in #(1)#, we get,sqrt

#dy/dx=(1/(2sqrtu))(3t^2)(cosv)(-4x)# #=-6xt^2cosv/sqrtu={-6xsin^2(2-x^2)cos(2-x^2)}/{sqrt(sin^3(2-x^2)}#=#-6xsqrtsin(2-x^2)cos(2-x^2).#

Charles Anctil · Answer

undefined To differentiate $ f(x) = \sqrt{\sin^3(2-x^2)} $ using the chain rule:

1. Identify the outer function $ g(x) = \sqrt{x} $ and the inner function $ h(x) = \sin^3(2-x^2) $.
2. Differentiate the outer function with respect to its argument: $ g'(x) = \frac{1}{2\sqrt{x}} $.
3. Differentiate the inner function with respect to $ x $:
   $ h'(x) = \frac{d}{dx}(\sin^3(2-x^2)) $.
4. Apply the chain rule: 
   $ \frac{d}{dx}(\sqrt{\sin^3(2-x^2)}) = g'(h(x)) \cdot h'(x) $.
5. Substitute the derivatives: 
   $ \frac{d}{dx}(\sqrt{\sin^3(2-x^2)}) = \frac{1}{2\sqrt{h(x)}} \cdot \frac{d}{dx}(\sin^3(2-x^2)) $.
6. Compute $ h'(x) $ using the chain rule and derivative of $ \sin^3(2-x^2) $:
   $ h'(x) = -3\sin^2(2-x^2) \cdot \cos(2-x^2) \cdot (-2x) $.
7. Simplify and combine terms to get the final result.