Find the derivative: f(x)=(3^((4x^5)+x-8))*(log2(6x+3)) I'm not sure that my answer is correct. Thank's!
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You should use the product rule and the chain rule to differentiate the function with respect to x, such that:
`f'(x) = (3^((4x^5)+x-8))'*(log_2(6x+3)) + 3^((4x^5)+x-8))*(log_2(6x+3))'`
You need to convert the given logarithm to natural logarithm such that:
`log_2(6x+3) = (ln(6x+3))/(ln 2)`
`f'(x) = (3^((4x^5)+x-8)*ln 3*(4x^5+x-8)'*(log_2(6x+3)) + 3^((4x^5)+x-8))*(1/(ln 2))*((6x+3)')/(6x+3)`
`f'(x) = (20x^4+1)*ln 3(3^((4x^5)+x-8)*((ln(6x+3))/(ln 2)) + 3^((4x^5)+x-8))*(1/(ln 2))*(6)/(6x+3)`
You should factor out `(3^((4x^5)+x-8))/(ln 2)` such that:
`f'(x) = (3^((4x^5)+x-8))/(ln 2)((20x^4+1)*ln 3*ln(6x+3) + 6/(6x+3))`
Hence, evaluating the given derivative of the function yields `f'(x) = (3^((4x^5)+x-8))/(ln 2)((20x^4+1)*ln 3*ln(6x+3) + 6/(6x+3)).`
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