Can you derive this plz?

Question

can you derive and give me the answer of h(t)= t^(-9t+5) Using ln.

Answer 1

In order to solve this, you are correct, you must use the natural log function (ln), followed by the log rules, followed by implicit differentiation.

Your first step is to take the ln of both sides.

ln ([h(t)] = ln [ t^(-9t + 5) ]

Remember the log rule that
log[base b](a^c) = c * log [base b](a), which essentially says whenever we have a power inside of a log, we can remove that power outside of the log and tack it on the front to multiply.

We use this log rule on the right hand side.

ln ([h(t)] = (-9t + 5) ln [t]

Now, we differentiate implicitly. Keep in mind that we'll be using the chain rule on the left hand side, and the product rule on the right hand side.

(1/[h(t)]) (h'(t)) = (-9) ln(t) + (-9t + 5)[1/t]

If we multiply both sides by h(t), we would eliminate the fraction on the left hand side and get

h'(t) = {h(t)} [ (-9) ln(t) + (-9t + 5)[1/t] ]

And now, we can actually replace h(t), to get

h'(t) = t^(-9t + 5) [ (-9) ln(t) + (-9t + 5)[1/t] ]

An exam question would usually not asking you to go further than this, but in theory we can simplify this by putting it under a common denominator, combining like terms, perhaps combining terms with exponents, and so forth.

Answer 2

rearrange:

log_t(h)=-9t+5
ln(h)/ln(t)=-9t+5

implicit diff:

ln(t)/h*(dh/dt)-ln(h)/t= -9ln(t)^2
ln(t)/h*(dh/dt)= ln(h)/t-9ln(t)^2
dh/dt= (ln(h)/t-9ln(t)^2)/(ln(t)/h)
dh/dt= h(ln(h)/t-9ln(t)^2)/ln(t)

sub h back

dh/dt= [t^(-9t+5)]((-9t+5)ln(t)/t - 9ln(t)^2) / ln(t)
dh/dt= [t^(-9t+5)]((-9t+5)/t - 9ln(t))
dh/dt= (5/t - 9ln(t) -9) t^(-9t+5)