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

Answer 1

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

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#.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

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

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))#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To differentiate ( f(x) = \cos(\sqrt{\cos(x^2)}) ) using the chain rule, follow these steps:

  1. Identify the outer function and inner function.

    • Outer function: ( \cos(x) )
    • Inner function: ( \sqrt{\cos(x^2)} )
  2. Find the derivative of the outer function with respect to its inner function.

    • ( \frac{d}{du} \cos(u) = -\sin(u) ), where ( u = \sqrt{\cos(x^2)} )
  3. Find the derivative of the inner function with respect to ( x ).

    • ( \frac{d}{dx} \sqrt{\cos(x^2)} = \frac{-\sin(x^2)}{2\sqrt{\cos(x^2)}} )
  4. Apply the chain rule:

    • ( \frac{d}{dx} \cos(\sqrt{\cos(x^2)}) = -\sin(\sqrt{\cos(x^2)}) \times \frac{-\sin(x^2)}{2\sqrt{\cos(x^2)}} )
  5. Simplify the expression if needed.

So, the derivative of ( f(x) = \cos(\sqrt{\cos(x^2)}) ) using the chain rule is: [ -\sin(\sqrt{\cos(x^2)}) \times \frac{-\sin(x^2)}{2\sqrt{\cos(x^2)}} ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7