find the derivative of 5 sinh(ln t). For this I tried so many different ways and I continually get it wrong.

Question

The function is 5 sinh(ln t). In all my methods I assumed 5 to be a constant, and pulled it out.

for the first try I said

5 cosh (ln(t))*1/t

then I tried using and replacing the e's in the hyperbolic function (my teacher doesn't recommend this for this problem) but I got

5((t-1/t)/2) and I got it wrong.

Then I tried rephrasing my first attempt because my teacher said parenthesis placement is super important. I got 5(cosh(ln(t)*1/t) and I still got it wrong.

On my last try I played with the e's again and I got 5/2 (1+ 1/t^2) and I also got it wrong.

I abandoned the e's however I am still not understanding how to take this derivative. I know the constant rule applies, but I just don't know how to find this correctly. Please help me.

Answer 1

Sinhx = (1/2)(e^x - e^-x)

 

So, sinh(lnt) = (1/2)(e^lnt - e^-lnt)

 

                  = (1/2)(t - e^{ln(t^-1)})

 

                  = (1/2)(t - t^-1)

 

If f(t) = 5sinh(lnt) = (5/2)(t - t^-1), then f'(t) = (5/2)(1 + 1/t²)

 

                                                              = (5(t² + 1))/(2t²)

 

 

 

Answer 2

Phil,

 

Let us take this step by step.  We need to apply the chain rule.  Your function is of the form

 

f(t) = 5 sinh(u(t)) 

 

u(t) = ln t

 

Then by the chain rule,

 

df/dt = df/du * du/dt

 

d/du (sinh u) = cosh u

 

d/dt (ln t) = 1/t

 

Then

 

d/dt (sinh (ln t)) = [cosh(ln t)] * 1/t

 

With the 5 in front,

 

f'(t) = 5 [cosh(ln t)] * 1/t

 

It looks like your first answer is correct.

 

If we put ln t inside the cosh, we get

 

cosh(ln t) = (1/2) * (e^{ln t} + e^{-ln t}) = (1/2) * (t + 1/t)

 

Then

 

f'(t) = 5 [(1/2) * (t + 1/t)] * 1/t = (5/2) * (1 + 1/t²)

 

It looks like your last answer is correct.  Did you enter the answer on-line?  Is there a strict parenthesis format that you might have missed?  Could the computer have interpreted your answer as 5 / [2(1 + 1/t²)] ?