# Derivatives and Integrals of Logarithms

• Mar 3rd 2009, 05:10 PM
sesshomaru
Derivatives and Integrals of Logarithms
I really need someone to atleast give me the answers to some of these problems because I have no clue as how to do half of them they're so much harder than the problems in the book and I have a test tomarrow (Doh)
1. Y=log5(3x-7)
2. y=9^2t
3. y=(x^2)(e^-2/x)
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• Mar 3rd 2009, 05:14 PM
Nacho
what you want? integrate those functions? o find your derivatives?

P.D: the last function is $\displaystyle y=x^2 \cdot e^{\frac{-2}{x}}$ ?
• Mar 3rd 2009, 05:17 PM
sesshomaru
Sorry about not being clear but for those I need the derivatives and the last one is

$\displaystyle e^(-2/x)$

If you cant tell the -2/x is the power
• Mar 3rd 2009, 05:18 PM
sesshomaru
Double post sorry
• Mar 3rd 2009, 05:27 PM
Nacho
do you know base change in logarithms?

1-. $\displaystyle y = \log _5 \left( {3x - 7} \right) = \frac{{\ln \left( {3x - 7} \right)}} {{\ln 5}}$

Whithot to forget the chain rule:

$\displaystyle y' = \frac{1} {{\ln 5}} \cdot \frac{1} {{3x - 7}} \cdot 3$

2-. You must know the identity: $\displaystyle a=e^{\ln a}$

$\displaystyle y = 9^{2t} = e^{2t\ln 9} \Rightarrow y' = 2\ln 9 \cdot e^{2\ln 9 \cdot t}$

Can you do the 3?
• Mar 3rd 2009, 05:28 PM
Mentia
Im sure you know that the derivative of the natural log of x is 1/x, i.e.:

$\displaystyle \frac{ d}{dx } ln(x) = \frac{1 }{x }$,

so you have a base 5 log in your first question. You should convert that to log base e so that you can do the derivative:

$\displaystyle { log}_{5 } (x) = \frac{ ln(x)}{ln(5) }$

You can get it from here I'm sure.
• Mar 3rd 2009, 05:48 PM
sesshomaru
Quote:

Originally Posted by Nacho
do you know base change in logarithms?

1-. $\displaystyle y = \log _5 \left( {3x - 7} \right) = \frac{{\ln \left( {3x - 7} \right)}} {{\ln 5}}$

Whithot to forget the chain rule:

$\displaystyle y' = \frac{1} {{\ln 5}} \cdot \frac{1} {{3x - 7}} \cdot 3$

2-. You must know the identity: $\displaystyle a=e^{\ln a}$

$\displaystyle y = 9^{2t} = e^{2t\ln 9} \Rightarrow y' = 2\ln 9 \cdot e^{2\ln 9 \cdot t}$

Can you do the 3?

Whats the answer to the first one? Shouldnt the Ln5 go away since it's a constant?

For the second one does the $\displaystyle e^{2\ln 9 \cdot t}$ need to be in the answer?

I have alot of questions that I need help with by the way im really struggling in this class and praying to pass with a C.
• Mar 3rd 2009, 06:04 PM
Nacho
I try explain you, but my english is a little basic ok?

In you first question, the constant, multiplying the variable, keeps

for example: $\displaystyle y = 3x = x + x + x \Rightarrow y' = 1 + 1 + 1 = 3$ but is more easye leave outside the constant

but is diferent to derivate a constant! which always worths zero$\displaystyle y = 3x + 4 \Rightarrow y' = 3 \cdot 1 + 0$

To second question: yes, you note $\displaystyle 2 \ln 9$ is a constan, then is same to derivate $\displaystyle y=e^c$ your derivative is $\displaystyle c \cdot e^c$

• Mar 3rd 2009, 06:14 PM
sesshomaru
Ok here's another one its an integral

$\displaystyle dx/(-x^2+4x-1)^(1/2)$

The the (-x^2 +4X-1) is under a square root. Dont worry your English is fine and thanks for helping me out.
• Mar 3rd 2009, 06:29 PM
Nacho
When I see this integral, my intuition say that is an integral of the form: $\displaystyle \int {\frac{1} {{\sqrt {1 - y^2 } }}dy}$ whic is easy to calculate

Then you note:

$\displaystyle \begin{gathered} - x^2 + 4x - 1 = - x^2 + 4x - 1 + 4 - 4 = - x^2 + 4x - 4 + 3 \hfill \\ = - \left( {x^2 - 4x + 4} \right) + 3 = 3 - \left( {x - 2} \right)^2 = 3\left\{ {1 - \frac{{\left( {x - 2} \right)^2 }} {3}} \right\} \hfill \\ \end{gathered}$

Then:

$\displaystyle \int {\frac{1} {{\sqrt { - x^2 + 4x - 1} }}} dx = \int {\frac{1} {{\sqrt {3\left\{ {1 - \frac{{\left( {x - 2} \right)^2 }} {3}} \right\}} }}} dx = \frac{1} {{\sqrt 3 }}\int {\frac{1} {{\sqrt {1 - \left( {\frac{{x - 2}} {{\sqrt 3 }}} \right)^2 } }}dx}$

Now the substitution is in all books:

$\displaystyle \begin{gathered} \sin z = \frac{{x - 2}} {{\sqrt 3 }} \Rightarrow \cos zdz = \frac{1} {{\sqrt 3 }}dx \hfill \\ \hfill \\ \frac{1} {{\sqrt 3 }}\int {\frac{1} {{\sqrt {1 - \sin ^2 z} }} \cdot \sqrt 3 \cos z} dz = \int {\frac{{\cos z}} {{\cos z}}dz} = z + C \hfill \\ \end{gathered}$

restituting:

$\displaystyle \int {\frac{1} {{\sqrt { - x^2 + 4x - 1} }}} dx = \arcsin \left( {\frac{{x - 2}} {{\sqrt 3 }}} \right) + C$