How do you calculate f''(1) and f'''(1) given #f(x)= 3 6(x1) + 4/(2!) (x1)^2  5/(3!) (x1)^3 + 1/(4!) (x1)^4#?
The Taylor series for
So
If you prefer , you could differentiate to get
So
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To calculate ( f''(1) ) and ( f'''(1) ), we need to find the second and third derivatives of the function ( f(x) ), and then evaluate them at ( x = 1 ).

Find ( f'(x) ): [ f'(x) = 6 + 4 \cdot \frac{2!}{2!}(x  1)^1  5 \cdot \frac{3!}{3!}(x  1)^2 + 1 \cdot \frac{4!}{4!}(x  1)^3 ] Simplify: [ f'(x) = 6 + 4(x  1)  5(x  1)^2 + (x  1)^3 ]

Find ( f''(x) ): [ f''(x) = 4  10(x  1) + 3(x  1)^2 ]

Find ( f'''(x) ): [ f'''(x) = 10 + 6(x  1) ]

Evaluate at ( x = 1 ): [ f''(1) = 4  10(1  1) + 3(1  1)^2 = 4 ] [ f'''(1) = 10 + 6(1  1) = 10 ]
Therefore, ( f''(1) = 4 ) and ( f'''(1) = 10 ).
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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.
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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