I understand if you refuse to give me the exact answer but I would appreciate help toward the answer. The answer I computed was f '(x)=(2e^2x)(4^2x)+(e^2x)[(4^2x)(2)ln(4)… I haven't simplified it but is this correct?
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f(x) = (e^2x)(4^2x)
f'(x) = (2e^2x)(4^2x) + (e^2x)(4^2x)(2)ln(4) => Yes. your answer is right!
Or another method:
y = (e^2x)(4^2x)
y = (e^X)(4^X) .... where X = 2x => dX/dx = 2
dy/dx = dy/dX * dX/dx = d[(e^X)(4^X)]/dX * dX/dx = [(e^X)*(4^X) + (e^X)*(4^X)*ln(4)] * 2
dy/dx = 2(e^X)*(4^X)[1 + ln(4)]
f'(x) = (2e^2x)(4^2x) + (e^2x)(4^2x)(2)ln(4) => Yes. your answer is right!
Or another method:
y = (e^2x)(4^2x)
y = (e^X)(4^X) .... where X = 2x => dX/dx = 2
dy/dx = dy/dX * dX/dx = d[(e^X)(4^X)]/dX * dX/dx = [(e^X)*(4^X) + (e^X)*(4^X)*ln(4)] * 2
dy/dx = 2(e^X)*(4^X)[1 + ln(4)]