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

1 Answer
Jun 23, 2016

Answer:

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

Explanation:

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).#