Differentiating this problem $\frac{2t^{3/2}}{\ln(2t^{3/2}+1)}$

Question

How does one differentiate the function $$y(t)=\frac{2t^{3/2}}{\ln(2t^{3/2}+1)}.$$

I am still tying to understand MathJaX and not sure what is wrong with the expression. Anyways,

How do I start/process solving this? Do i take the ln of both side? If so I get the log of the top - the log of the bottom. which is the log of a log? If I do the quotient rule right away, i get the log expression in the bottom squared. Help please?

Answer 1

In this case the quotient rule is probably the best option. The symmetry of the $2 t^{2/3}$ in the top and bottom makes me suspect that some things might end up canceling out.

Answer 2

I'm pretty sure this answer will be satisfactory. Usually, you aren't asked to simplify more than this.

Answer 3

Getting the square of the logarithm in the denominator is a fact. You have tools for differentiating, so apply them $$y(t)=\frac{2t^{3/2}}{\ln(2t^{3/2}+1)}\\ y'(t)=\frac{3t^{1/2}\ln(2t^{3/2}+1)-2t^{3/2}\frac1{2t^{3/2}+1}3t^{1/2}}{((\ln(2t^{3/2}+1))^2}$$ which you can simplify if you want by distributing out the $3t^{1/2}$, but that is in the eye of the beholder.