Applying the rules for the constant, logarithm and exponents, the derivative of the given function is of:
f'(x) = 3/x - 6x² - 8/(x^5).
<h3>What is the derivative of the function?</h3>
The function is given by:
f(x) = √2^3 + ln(x^3) - 2x^3 - log_3(729) - 2/x^4
The derivative of a sum is the sum of the derivatives, hence:
f'(x) = (√2^3)' + (ln(x^3))' - (2x^3)' - (log_3(729))' - (2/x^4)'
The derivative of a constant is of zero, hence:
Hence:
f'(x) = (ln(x^3))' - (2x^3)' - (2/x^4)'.
The derivative of the logarithm is given by:
ln(h(x))' = (1/h(x)) x h'(x).
Hence:
(ln(x^3))' = 3x²/x³ = 3/x.
The derivative of the power of x is given by:
[x^n]' = nx^(n-1).
Hence:
- (2/x^4)' = (2x^(-4))' = -8x^(-5) = -8/(x^5).
Hence the derivative of the function is given by:
f'(x) = 3/x - 6x² - 8/(x^5).
More can be learned about derivatives at brainly.com/question/23819325
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