# Derivative of exponential problem.

#### spoc21

Hi, Im having difficulties solving the following two problems

[FONT=&quot]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:
s(t) =
$$\displaystyle 1/(3t+1)$$..

Im really confused here

For b) I'm lost on how to find the implicity differentiation of 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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#### VonNemo19

MHF Hall of Honor
Hi, Im having difficulties solving the following two problems

[FONT=&quot]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:
s(t) =
$$\displaystyle 1/(3t+1)$$..

Im really confused here

For b) I'm lost on how to find the implicity differentiation of 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
[FONT=&quot]

[/FONT]
For a) Since $$\displaystyle \frac{d}{dx}[\ln{u}]=\frac{1}{u}u'$$ we have $$\displaystyle u=3t^2+t$$, so that $$\displaystyle \frac{d}{dt}[\ln{u}]=\frac{1}{3t^2+t}\cdot(6t+1)$$

#### HallsofIvy

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

[FONT=&quot]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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