Derivatives of Log and Exponential Functions

AI Thread Summary
The discussion revolves around solving derivatives of various logarithmic and exponential functions. Participants check each other's work on specific problems, including derivatives of functions like y=e^sin3x and y=ln(x^2/(2x+5)^3). There are also inquiries about radioactive decay functions, where the initial amount and decay rate are calculated, and participants clarify how to isolate dy/dx in complex equations. Additionally, there is a focus on solving logarithmic equations, with guidance provided on using properties of logarithms to find solutions. Overall, the thread emphasizes collaborative problem-solving in calculus.
majinknight
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Hi i was wondering if someone could check my work for these couple questions:

Find dy/dx (do not simplify)

a)y=e^sin3x
dy/dx= e^sin3x (cos3x)(3)
=3cos3xe^sin3x

b)y=5^(square rootx) x^2
dy/dx=x5^(square rootx) (xln5+2)

c)y=ln(x^2 / (2x+5)^3 )
y=lnx^2 - ln(2x+5)^3
dy/dx= 2x/x^2 - 6(2x+5)^2 / (2x=5)^3
=2/x - 6/(2x+5)

d)y=4log base 2 (square rootx+1)
y=4logbase2 (x+1)^-1
dy/dx= 4/lnbase2(square rootx+1)

e)y=ln[x^2 - e^x / x^2 +e^x]
y= ln(x^2 -e^x) - ln(x^2 +e^x)
dy/dx= [2x -e^x / x^2 - e^x] - [2x +e^x / x^2+e^x]

f)e^x^2 multiplied by y^3=x (isolate dy/dx)
I am unsure of how to do this one, i have never seen one like this before, could someone show what i would do.

g)Use logarthimic differentiation to find dy/dx if
y=[e^x cosx / (square root x)]^5

lny=5xlne +5lncosx -5/2lnx
dy/dx= y[5x-5tanx-5/2x]
dy/dx= =[e^x cosx / (square root x)]^5 [5x-5tanx-5/2x]

h)A radioactive substance decays in such a way that the amount in grams present at time t years is given by A(t)=100e^-0.2t
i)What is the initial amount of radioactive material, A(0)?
A(0)= 100
ii)Find the rate of decay function, A'(t).
A'(t)=-20e^-0.2t
iii)How mucgh radio active material is present when t-50 years? How fast is the material decaying at this time?
A(50)=100e^-10
=0.0045399
A'(50)=-20e^-10
=-0.0009079
iv)At what time t is one half of the original substance remaining? What is the decay rate at this time?
I am not sure what to do here also if someone could show me how id really appreciate it.
*For this question and parts of the question i am not sure if some of my equations are correct as the numbers i am getting do not seem like it should be what they are.

j)solve the logarithmic equation logbase3(x-3) + logbase3(x) = logbase3(4)
I am also not sure what to do for this question.

Thanks in advanced for the help! If someone could show me how to do those few questions and check my work on those ones because my textbook does not show what the answers are or how to do some of them.
 
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majinknight said:
Hi i was wondering if someone could check my work for these couple questions:
Find dy/dx (do not simplify)

a)y=e^sin3x
dy/dx= e^sin3x (cos3x)(3)
=3cos3xe^sin3x

This is going to be a hell of a long post. :-p This part is correct.

majinknight said:
b)y=5^(square rootx) x^2
dy/dx=x5^(square rootx) (xln5+2)

\ln y=\sqrt{x}\ln5 +2\ln x;\frac{d\ln y}{dx}=\frac{1}{y}\frac{dy}{dx}
So:\frac{dy}{dx}=y\frac{d\ln y}{dx}=(5^{\sqrt{x}}x^{2})(\frac{\ln 5}{2\sqrt{x}}+\frac{2}{x})=...

majinknight said:
c)y=ln(x^2 / (2x+5)^3 )
y=lnx^2 - ln(2x+5)^3
dy/dx= 2x/x^2 - 6(2x+5)^2 / (2x+5)^3
=2/x - 6/(2x+5)

Okay...

majinknight said:
d)y=4log base 2 (square rootx+1)
y=4logbase2 (x+1)^-1
dy/dx= 4/lnbase2(square rootx+1)

I don't understand where is the square root and what's that argument for the \log_{2}

majinknight said:
e)y=ln[x^2 - e^x / x^2 +e^x]
y= ln(x^2 -e^x) - ln(x^2 +e^x)
dy/dx= [2x -e^x / x^2 - e^x] - [2x +e^x / x^2+e^x]

I don't follow you.If u have:
y=\ln(x^{2}-\frac{e^{x}}{x^{2}}+e^{x}),then the decomposition u found is wrong.It's correct iff u have
y=\ln(\frac{x^{2}-e^{x}}{x^{2}+e^{x}})
Then your result would be correct.However,the fact that you're not writing it in 'tex' misleads me.

majinknight said:
f)e^x^2 multiplied by y^3=x (isolate dy/dx)
I am unsure of how to do this one, i have never seen one like this before, could someone show what i would do.

Here u got me completely lost.Is it
e^{x^{2}}y^{3}=x?If so,then take natural log.out of both sides and get:
x^{2}+3\ln y=\ln x.Differentiate wrt to 'x' to find
2x+\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}.From there extract the derivative and substitute 'y' by the expression foud from the first equation.

majinknight said:
g)Use logarthimic differentiation to find dy/dx if
y=[e^x cosx / (square root x)]^5

lny=5xlne +5lncosx -5/2lnx
dy/dx= y[5x-5tanx-5/2x]
dy/dx= =[e^x cosx / (square root x)]^5 [5x-5tanx-5/2x]

Okay.

majinknight said:
h)A radioactive substance decays in such a way that the amount in grams present at time t years is given by A(t)=100e^-0.2t
i)What is the initial amount of radioactive material, A(0)?
A(0)= 100
ii)Find the rate of decay function, A'(t).
A'(t)=-20e^-0.2t
iii)How mucgh radio active material is present when t-50 years? How fast is the material decaying at this time?
A(50)=100e^-10
=0.0045399
A'(50)=-20e^-10
=-0.0009079

So far so good.
majinknight said:
iv)At what time t is one half of the original substance remaining? What is the decay rate at this time?
I am not sure what to do here also if someone could show me how id really appreciate it.

You found that,initially,A(0)=100.The question is:find 't' from the eq.
A(t)=100e^{-0.2t}=50


majinknight said:
j)solve the logarithmic equation logbase3(x-3) + logbase3(x) = logbase3(4)
I am also not sure what to do for this question.

Raise both sides of the eq. to the power 3.Use the property
3^{x+y}=3^{x}3^{y}
to find the solution.

Daniel.
 
Ya some of it is confusing as i don't know how to use that tex thing. For question d) the root is over x+1. And for question e) the equation is the second one you posted which as you said my answer is correct. f) is the equation you said so then i did what you said and this is the answer i get.

dy/dx= (cube root of x-e^x^2)(1/x -2x)
Is that correct?

And for h)iv)
50=100e^-0.2t
ln0.5=-0.2t
t=ln0.5/-0.2
so then:
A'(ln0.5/-0.2)=-20e^ln0.5
Would i leave my answer at that, and is that correct?

For question j) i still don't understand what you mean, could you please show me what the first line of the solution would be please. Thanks so much for the help!
 
For (j), he means this:

\log_b{xy} = \log_b{x}+\log_b{y}
 
majinknight said:
Ya some of it is confusing as i don't know how to use that tex thing. For question d) the root is over x+1. And for question e) the equation is the second one you posted which as you said my answer is correct. f) is the equation you said so then i did what you said and this is the answer i get.

dy/dx= (cube root of x-e^x^2)(1/x -2x)
Is that correct?

And for h)iv)
50=100e^-0.2t
ln0.5=-0.2t
t=ln0.5/-0.2
so then:
A'(ln0.5/-0.2)=-20e^ln0.5
Would i leave my answer at that, and is that correct?

For question j) i still don't understand what you mean, could you please show me what the first line of the solution would be please. Thanks so much for the help!

d)y=4\log_{2}\sqrt{x+1} \Rightarrow 2^{y}=2^{4\log_{2}\sqrt{x+1}}=(2^{\log_{2}\sqrt{x+1}})^{4}=(\sqrt{x+1})^{4}=(x+1)^{2}
Take '\ln' from both sides of the eq.to find:
y=\frac{2}{\ln 2}\ln(x+1)\Rightarrow \frac{dy}{dx}=\frac{2}{(\ln 2)(x+1)}


h)iv) t=5\ln 2 \Rightarrow A'(5\ln 2)=20e^{-0.2\cdot 5\cdot\ln 2}=\frac{20}{2}=10

It's not correct what u've written.Mass cannot be negative.

j)x(x-3)=4 \Rightarrow x=4
The solution x=-1 implies logarithm in the base "3" from a negative number...

Daniel.
 
Last edited:
For f):
\frac{dy}{dx}=y(\frac{1}{x}-2x)=(\sqrt[3]{x e^{-x^{2}}})(\frac{1}{x}-2x)


Daniel.
 
Alright thanks i understand now! Thanks so much.
 
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