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differentiate: y(t)=(1)/(t+4e^t)
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so i know well be using the product rule..but i just dont know what to do with it after i took the derivative
it's chain rule : y= (t + 4e^t)^-1 dy/dt = [-(t + 4e^t)^-2] x derivative of (t + 4e^t) dy/dt = [-(t + 4e^t)^-2] x (1 + 4e^t)
its a fraction 1 over t + 4e^t
\[[-(t + 4e^t)^-2] \times d(t + 4e^t)/dt\] \[dy/dt = [-(t + 4e^t)^-2] \times (1 + 4e^t)\]
aha .. so you differentiate the outer as it is u^-1 ,, then multiply it by the derivative of the inner (t + 4e^t) this is the chain rule yes I make 1 over t + 4e^t = (t + 4e^t)^-1
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did you understand ??
\[d/dx{\frac{1}{f(x)}=\frac{-f'(x)}{f^2(x)}}\] so you answer here is \[\frac{-(4e^x+1)}{(x+4e^x)^2}\]
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