Hi, I`m having difficulties solving the following two problems

[FONT="]a) s(t) = ln(3t2 + t) find the slope of the function at t = 2

b) [/FONT] Find dy/dx for the function xy2 + xln x = 4y for x > 0.

For a), Im completely lost on how to find the derivative of the s(t)

..would be just take the inverser of the function, using the derivative, so that:

That is the reciprocal, not the inverse function. The reciprocal is the **mulitplicative** inverse but when you are talking about functions, "inverse function" is completely different.

\(\displaystyle 1/(3t+1)\)..

First, it's not "3t+ 1", it is \(\displaystyle 3t^2+ 1\) so you would start with \(\displaystyle \frac{1}{3t^2+ 1}\) but then, by the chain rule, you have to multiply by the derivative of \(\displaystyle 3t^2+1\).

Im really confused here

For b) I'm lost on how to find the implicity differentiation of the exponential function.

\(\displaystyle xy^2 + xln x = 4y\)

Differentiate each part: \(\displaystyle (xy^2)'+ (x ln x)'= (4y)'\). Use the product rule for the first term: \(\displaystyle (xy^2)'= (x)'(y^2)+ (x)(y^2)'\) of course, \(\displaystyle (x)'= 1\) and using the chain rule,\(\displaystyle (y^2)'= (2y y')\) so \(\displaystyle (xy^2)'= y^2+ 2xyy'\).

Use the product rule for the second term: \(\displaystyle (x ln x)'= (x')(ln x)+ (x)(ln x)'\). Again \(\displaystyle (x)'= 1\) and \(\displaystyle (ln x)'= \frac{1}{x}\) so \(\displaystyle x ln x)'= ln x+ \frac{x}{x}= ln x+ 1\).

Finally, (4y)' is just 4y' so altogether we have

\(\displaystyle y^2+ 2xyy'+ ln x+ 1= 4y'\)

Solve that equation for y'.

By the way, these are

**not** "exponential functions". They are "logarithmic functions", the inverse (

**not** "reciprocal"!!!) function to the exponential function.

I've been trying to solve these problems over the entire weekend, but have had no luck, and would really appreciate any helful advice/tips on solving the problem.

Thanks

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