Question

If `y = sin^4x -cos^4x`, what is `dy/dx`?

Answer 1

Use the chain rule to differentiate both terms

Explanation:

The chain rule states `f'(g(x)) = f'(x)*g'(x)`
We can let `f(x) = x^4 and g(x) = sin(x)`, thus `f(g(x)) = sin^4(x)`
Similarly, we can let `h(x) = cos(x)` and so `f(h(x)) = cos^4(x)`

So then, `f'(g(x)) = 4(sin^3(x))*cos(x) = 4cos(x)sin^3(x)`

Similarly, `f'(h(x)) = 4cos^3(x)*(-sin(x)) = -4sin(x)cos^3(x)`

Thus the complete derivative becomes `4cos(x)sin^3(x) - -4sin(x)cos^3(x) = 4cos(x)sin^3(x) + 4sin(x)cos^3(x)`

The final answer can be simplified using trigonometric identities.

Answer 2

`dy/dx=4sin^3xcosx+4sinxcos^3x`

`=2sin(2x)`

Explanation:

`y=sin^4x-cos^4x`

Use the chain rule to differentiate:

The chain rule states that: `(f(g(x)))'=f'(g(x))*g'(x)`

`dy/dx=4sin^3x(cosx)-4cos^3x(-sinx)`

`dy/dx=4sin^3xcosx+4sinxcos^3x`

Use the following trigonometric identities to simplify:

`color(blue)(sin^2x+cos^2x=1)`
`color(red)(sin(2x)=2sinxcosx)`

`dy/dx=sinxcosx[4color(blue)(sin^2x)+4color(blue)(cos^2x)]`

`=sinxcosx[4(color(blue)(sin^2x+cos^2x))]`

`=4sinxcosx`

`=2(color(red)(2sinxcosx))`

`=2color(red)(sin(2x))`

Answer 3

`dy/dx= 2sin2x`

Explanation:

We notice that the function is a difference of squares.

Therefore,

`y= (sin^2x- cos^2x)(sin^2x + cos^2x)`

Use `sin^2x+ cos^2x = 1`:

`y = sin^2x - cos^2x`

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

Now use `cos2x= cos^2x- sin^2x`

`y = -cos2x`

Now by the chain rule, we have:

`dy/dx = 2sin(2x)`

This is because the derivative of `-cosa = a'sina`

Hopefully this helps!