What is the second derivative of #f(x)=cos(x)sinx#?
The function can be simplified as
Differentiating once:
Differentiating again:
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To find the second derivative of ( f(x) = \cos(x)  \sin(x) ), we first differentiate the function ( f(x) ) with respect to ( x ) twice.
First derivative: [ f'(x) = \frac{d}{dx}[\cos(x)]  \frac{d}{dx}[\sin(x)] ]
Using the chain rule and derivative of cosine and sine functions: [ f'(x) = (1)\sin(x)  \cos(x) ]
[ f'(x) = \sin(x)  \cos(x) ]
Now, taking the derivative of ( f'(x) ) with respect to ( x ) gives the second derivative: [ f''(x) = \frac{d}{dx}[\sin(x)  \cos(x)] ]
[ f''(x) = \frac{d}{dx}[\sin(x)]  \frac{d}{dx}[\cos(x)] ]
Using the derivative of sine and cosine functions: [ f''(x) = \cos(x) + \sin(x) ]
So, the second derivative of ( f(x) = \cos(x)  \sin(x) ) is ( f''(x) = \cos(x) + \sin(x) ).
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To find the second derivative of ( f(x) = \cos(x)  \sin(x) ), follow these steps:

Find the first derivative of ( f(x) ): [ f'(x) = \frac{d}{dx} [\cos(x)]  \frac{d}{dx} [\sin(x)] ]

Use the chain rule and derivative of cosine and sine functions: [ f'(x) = \sin(x)  \cos(x) ]

Simplify: [ f'(x) = \sin(x)  \cos(x) ]

Now, find the second derivative of ( f(x) ): [ f''(x) = \frac{d}{dx} [\sin(x)  \cos(x)] ]

Differentiate each term: [ f''(x) = \frac{d}{dx} [\sin(x)]  \frac{d}{dx} [\cos(x)] ]

Apply the derivative rules for sine and cosine functions: [ f''(x) = (\cos(x))  (\sin(x)) ]

Simplify: [ f''(x) = \cos(x) + \sin(x) ]
So, the second derivative of ( f(x) = \cos(x)  \sin(x) ) is ( f''(x) = \cos(x) + \sin(x) ).
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When evaluating a onesided 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 onesided 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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