How do you differentiate #f(x)=sin(4-x^2) # using the chain rule?

Question

1428 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-04T14:07:17

#f' (x)=-2x*cos (4-x^2)#

The solution :

The formula for differentiating #sin u# for any differentiable function #u# is

#d/dx(sin u)=cos u d/dx(u)#

The given: #f(x)=sin (4-x^2)#

#f' (x)=d/dx(sin (4-x^2))=cos (4-x^2) d/dx(4-x^2)#

#f' (x)=cos (4-x^2) *(-2x)#

#f' (x)=-2x*cos (4-x^2)#

God bless....I hope the explanation is useful.