How do you differentiate #f(x)=cos(sqrt((cosx^2))) # using the chain rule?

2 Answers
Nov 10, 2015

Answer:

#f'(x) = (sin(sqrt(cos x^2)))/( sqrt(cos x^2)) cdot ( sin x^2)cdot x#

Explanation:

Use progressively the chain rule which says :
#(d F(u(x))) / dx = F'(u(x)) cdot u'(x)#.

#f'(x) = cos'(sqrt(cos x^2)) cdot (d sqrt(cos x^2))/dx#
so,
#f'(x) = -sin(sqrt(cos x^2)) cdot 1/(2 sqrt(cos x^2)) cdot (d cos x^2)/dx#
because #cos' = -sin# and #(d\sqrt(X))/(dX) = 1/(2sqrt(X))#.
Therefore;
#f'(x) = -(sin(sqrt(cos x^2)))/(2 sqrt(cos x^2)) cdot (-( sin x^2)cdot (d x^2)/(dx))#
Finally,
#f'(x) = (sin(sqrt(cos x^2)))/( sqrt(cos x^2)) cdot ( sin x^2)cdot x#
because # (d x^2)/(dx)=2x#.

Nov 10, 2015

Answer:

#dy/dx = f^'(x) = (sin(sqrt(cos(x^2)))sin(x^2)x)/(sqrt(cos(x^2))#

Explanation:

I'll be using Leibniz notation because I think it's easier to understand more complicated chain rule applications with this.

The chain rule, in said notation goes as follows,

For #y = f(x)# such that #f(x) = g(u)# then #dy/dx = d/(du)(g(u))(du)/dx#,

Or, if we can say that a function #f(x)# is a simpler function #g(u)# for a defined relationship between #x# and #u#, we can derive #g(u)# wrt #u# and multiply by the derivative of #u#.

In this case we have

#y = cos(sqrt(cos(x^2)))#

By saying #sqrt(cos(x^2)) = u# we have

#dy/dx = d/(du)cos(u)(du)/dx#

Or

#dy/dx = -sin(u)(du)/dx#

Now, if we say that #cos(x^2) = v# we can say that

#dy/dx = -sin(u)d/(dv)(sqrt(v))(dv)/dx#

Or

#dy/dx = -sin(u)/(2sqrt(v))(dv)/dx = -sin(u)/(2u)(dv)/dx#

If we say #x^2 = w# we have

#dy/dx = -sin(u)/(2u)d/(dw)cos(w)(dw)/dx#

Or

#dy/dx = -sin(u)/(2u)(-sin(w))(dw)/dx = (sin(u)sin(w))/(2u)(dw)/dx#

Or, finally

#dy/dx = (sin(u)sin(w))/(2u)d/dx(x^2)#
#dy/dx = (sin(u)sin(w)x)/(u)#

And now we subsitute everything back to terms of #x#

#dy/dx = f^'(x) = (sin(sqrt(cos(x^2)))sin(x^2)x)/(sqrt(cos(x^2))#