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If f(x) = (x⁶-x⁴)⁵, then the 31st derivative of x=0 is?

A. 0
B. 31!
C. 5!
D. [infinity]

User Romhein
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1 Answer

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Final answer:

To find the 31st derivative of f(x) = (x⁶-x⁴)⁵ at x=0, the chain rule is applied repeatedly, resulting in a final answer of 0.

Step-by-step explanation:

To find the 31st derivative of f(x) = (x⁶-x⁴)⁵ at x=0, we can use the chain rule. The derivative of (x⁶-x⁴)⁵ is given by:

f'(x) = 5(x⁶-x⁴)⁴(6x⁵-4x³)

Continuing this process, we can find the 31st derivative by repeatedly applying the chain rule:

f''(x) = 20(x⁶-x⁴)³(36x¹⁰-12x⁸)
...
f⁽³⁰⁾(x) = 5(6x¹⁸-4x¹⁶)
f⁽³¹⁾(x) = 0

Therefore, the 31st derivative of f(x) at x=0 is 0. Answer choice A is correct.

User Neistow
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