What is the #n^(th)# derivative of #sin^2x#?

Answer 1

# f^((n)) sin^2x = { (sin^2x,n=0), ((-1)^(n/2+1) \ 2^(n-1) \ cos 2x,n gt 0 " even"), ((-1)^((n+1)/2+1) \ 2^(n-1) \ sin 2x,n gt 0 " odd") :} #

We have:

# f(x) = sin^2x #
Differentiating once wrt #x# (using the chain rule), we get the first derivative:
# f'(x) = 2sinxcosx #

At first glance we may suspect that to gain further derivatives we will require the product rule and their form will become increasingly more complex. However we note that:

# sin 2A -= 2sinAcosA #

Allowing us to write the first derivative as:

# f'(x) = sin2x #

So differentiating further times we get:

# f^((2))(x) = 2cos2x # # f^((3))(x) = -2^2sin2x# # f^((4))(x) = -2^3cos2x# # f^((5))(x) = 2^4sin2x# # f^((6))(x) = 2^5cos2x# # f^((7))(x) = -2^6sin2x# # f^((8))(x) = -2^7cos2x# # vdots #
And a clear pattern is now forming, and the #n^(th)# derivative is:
# f^((n)) sin^2x = { (sin^2x,n=0), ((-1)^(n/2+1) \ 2^(n-1) \ cos 2x,n gt 0 " even"), ((-1)^((n+1)/2+1) \ 2^(n-1) \ sin 2x,n gt 0 " odd") :} #
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Answer 2

The n^(th) derivative of sin^2(x) is:

[ \frac{{d^n}}{{dx^n}} (\sin^2(x)) = \sin^2(x) + 2n\sin(x)\cos(x) ]

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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.

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