Derivatives of Log and Exponential Functions

[majinknight]

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.

 

[dextercioby]

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. :tongue2: This part is correct.

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

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...

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 $\log_{2}$

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.

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.

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.

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.

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.
$$A(t)=100e^{-0.2t}=50$$

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.

 

[majinknight]

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!

 

[nolachrymose]

For (j), he means this:

$$\log_b{xy} = \log_b{x}+\log_b{y}$$

 

[dextercioby]

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.

 

[dextercioby]

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


Daniel.

 

[majinknight]

Alright thanks i understand now! Thanks so much.