What is the derivative of this function #f(x) = sin (1/x^2)#?

Answer 1

#(df(x))/dx= (-2cos(1/x^2))/x^3#

This is a simple chain rule problem. It is a little easier if we write the equation as: #f(x) = sin(x^-2)#
This reminds us that #1/x^2# can be differentiated the same way as any polynomial, by dropping the exponent and and reducing it by one.
The application of the chain rule looks like: #d/dx sin(x^-2) = cos(x^-2)(d/dx x^-2)# #= cos(x^-2)(-2x^-3)# #= (-2cos(1/x^2))/x^3#
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Answer 2

The derivative of the function ( f(x) = \sin(1/x^2) ) is:

[ f'(x) = -\frac{2\cos(1/x^2)}{x^3} ]

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Answer 3

To find the derivative of the function ( f(x) = \sin\left(\frac{1}{x^2}\right) ), we can use the chain rule.

Let ( u = \frac{1}{x^2} ), then ( \frac{du}{dx} = -\frac{2}{x^3} ).

Now, applying the chain rule, we have:

[ \frac{df}{dx} = \cos\left(\frac{1}{x^2}\right) \cdot \frac{d}{dx}\left(\frac{1}{x^2}\right) ]

[ = \cos\left(\frac{1}{x^2}\right) \cdot \left(-\frac{2}{x^3}\right) ]

[ = -\frac{2 \cos\left(\frac{1}{x^2}\right)}{x^3} ]

Therefore, the derivative of the function ( f(x) = \sin\left(\frac{1}{x^2}\right) ) is ( -\frac{2 \cos\left(\frac{1}{x^2}\right)}{x^3} ).

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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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