1. What is the derivative of f(x) = cos (x4)?

Question

1. What is the derivative of f(x) = cos (x⁴)?

2. What is the derivative of g(x) = (cos x)⁴?

3. Use implicit differentiation to find y’ of x² + y² = 8 at (2,2) and at (-2,2) (it may be helpful to look at the graph)

Answer 1

To determine how to use the chain rule, we need to figure out what is the "inner" function and what is the "outer" function. Usually, this is clear with polynomials like (3x² +2)⁴ where the inner function is the part in parenthesis and the outer is the 4th power. 

 

With trig functions, the inner is always the part that is in the parenthesis. So let's look at your given problems now.

 

1) f(x) = cos(x⁴)

 

The inner is h=x⁴ and the outer is cos(h) [I'm giving it a different variable so that it's more clear what to do]. We first take the derivative of the outer function. The derivative of cos is (-sin)

 

-sin(h)

 

Now we use the chain rule to multiply by the derivative of the inner function, which is

 

d/dx[h] = [dh/dx] = 4x³    using the power rule.

 

The full derivative is 

 

d/dx [cos(h)] * d/dx[h]

-sin(x⁴) * 4x³

 

-4x³sin(x⁴)

 

2) This time, the inner function is cosx. We use the same general ideal. First, we find the derivative of the outer function, which is h⁴

 

4*h³

 

Then multiply by the derivative of the inner function and do the same substitution.

 

4(cosx)³ * (-sinx)

-4sin(x) * [cos(x)]^{3  you will often see this written as}

 

-4sin(x)*cos³(x)

 

3) Implicit differentiation is much more complicated to explain in text format, so I will just show the steps. Essentially it is a bunch of power and chain rules written out very literally.

 

2x¹*d/dx[x] + 2y¹*d/dx[y] = d/dx[8]

 

2x*(1) + 2y*[dy/dx] = 0      note that [dy/dx] is the same thing as y', but it looks confusing in text here with the exponents

 

2x + 2y * [dy/dx] = 0

 

Solve for y', or [dy/dx]

 

[dy/dx] = -2x/(2y)

            = -x/y

 

It now says to find [dy/dx] at 2 pair of x,y values, so just plug those in.

 

a) [dy/dx] = -(2)/2 = -1

b) [dy/dx] = -(-2)/2 = 1

 

The hint about looking at the graph is that the given equation is that of a circle with a center at the origin (and radius sqrt(8) ). So, for positive values of X, the slope (a.k.a. derivative) should be negative and for negative values of X, the slope should be positive when both Y values are also positive.

 

I hope this helps! Please ask any questions if you have them.