Question

What is the derivative of `f(x)=cos^2x-cosx^2dx`?

Answer 1

`f'(x) = 2xsin(x^2) - sin2x`

Explanation:

I assume the `dx` is an error and that we have:

`f(x) = cos^2x-cos(x^2)`

We differentiate this by using the chain rule on both components separately;

For `cos^2x`; Let `A=cos^2x`, Then:

# { ("Let",u=cosx, => (du)/dx=-sinx), ("Then",A=u^2, => (dA)/(du)=2u ) :}#

Using the chain rule `(dA)/dx=((dA)/(du))((du)/dx)` we get:

`(dA)/dx = (2u)(-sinx)`
`" " = (2cosx)(-sinx)`
`" " = -sin2x`

And, for `cos(x^2)`; Let `B=cos(x^2)`, Then:

# { ("Let",u=x^2, => (du)/dx=2x), ("Then",B=cosu, => (dB)/(du)=-sinu ) :}#

Using the chain rule `(dB)/dx=((dB)/(du))((du)/dx)` we get:

`(dB)/dx = (-sinu)(2x)`
`" " = -2xsin(x^2)`

And combining the results we get:

`f'(x) = 2xsin(x^2) - sin2x`

Answer 2

`f(cos ^2 (x)-xsin (2x )+ 3dx^2sin(dx^3 )`

Explanation:

`d/(dx)(f(x) = cos^2x - cos x^2dx)`

Apply the sum/difference rule

= `d/(dx)(fx = cos^2(x) - d/(dx)(cos (x^2dx))`

Treat f, d as constants

`d/(dx)(fx = cos^2(x))`

Take the constant out of the brackets

`f*d/(dx) = (xcos^2(x))`

Appy the product rule f*g = f+g + f+g

where f = x g = cos^2

`f(d/(dx) (x)cos^2(x) + d/(dx)(cos^2(x))x)`

Keep the f and #cos^2(x) separate and solve rest of the equation

`d/dx * x = x/x = 1`

d/dx(cos^2x)

Apply chain rule

`(df(u))/(dx) = (df)/(du) * (du)/(dx)`

Let cos(x) be u

= `d/(du )(u^2) * d/(dx)(cos(x))`

Divide this equation into to and solve the first part

= `d/(du) (u^2) = 2u`

Then solve the second part

`d/(dx)(cos(x))`

= `-sin(x)`

Now combine both the equations

`2u(-sin(x))`

substitute back cos(x)

`2cos(x) (-sin(x))`

= `-2cos(x)(sin(x)`

Bring back the f and `cos^2(x)`

= `f(1*cos^2(x) + (-2cos(x)sin(x))x)`

=`f(-xsin(2x) + 1*cos^2(x))`

= `f(+1*cos^2(x) - xsin(2x)`

Remember this equation `d/(dx) (cos(x^2dx))` which we avoided

Now time to solve this

Apply the chain rule and let `x^2dx` be u

=`d/(du)(cos(u))* d/dx(x^2dx)`

Again separate the equation into two.

`d/(du)(cos(u)) = -sin(u)`

Now solve for the second part

`d/(dx)(x^2dx) =dd/dx(x^2x)`

= `3dx^2`

= `-sin(u) * 3dx^2`

Now substitute back `x^2dx` in u

= `-sin(x^2dx)*3dx^2`

=`-3dx^2sin(dx^3)`

Now combine all the equations

`f(cos ^2 (x)-xsin (2x )-(-3dx^2sin(dx^3 )`

`f(cos ^2 (x)-xsin (2x )+ 3dx^2sin(dx^3 )`