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find f'(x) of f(x)= 4x + 5sin^4x
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f'(x) = 4 + 20cos4x chain rule used to differentiate the second term
i dont see how that is the answer
well derivative of 4x = 4 right?
something would be to the third power though, wouldnt it?
that x is not in the power \[f(x) = 4x +5\sin^4 x\] i dont know if that makes a difference
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standard result: if f(x) = ax^n then f'(x) = an x^(n-1) so for 4x = 4x^1 f'(x) = 4*1 x^(1-1) = 4 x^0 = 4
oh sorry i misread the second term - i took it to be 5sin4x
so would it be 4+20sinx^3 * cosx?
ok derivative of second term is 4 sin^3 x * 5 cos x = 20cosxsin^3 x 4 + 20cosxsin^3x is your answer
yes you are right
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