logarithmic derivatives

**bassman111** · October 18th 2009, 04:01 PM

I've gotten some excellent help on these forums today. A HUGE thank you to everyone who assisted.

3 more questions giving me issues. If you could at least help get me started I would appreciate that...no need to do all my homework for me lol

Find the derivative of the following functions-

f(t)=tan^-1(e^4t-2)tan(e^t+1)

g(z)=ln(z^2)-5ze^(2z-1)

Given e^(x^2)+y^6=3xtan(y), find y'

I'm not really sure where to start on these

**Scott H** · October 18th 2009, 04:14 PM

Do you remember the Power Rule, the Product Rule, and the Chain Rule? Here are two additional formulas that will help:

$\frac{d}{dx}\ln x=\frac{1}{x}\;\;\;\;\;\;\;\;\;\;\frac{d}{dx}\tan^ {-1}x=\frac{1}{1+x^2}$

For the final problem, we may differentiate both sides and express $y'$ in terms of $x$ and $y$ .

**adkinsjr** · October 18th 2009, 04:17 PM

Originally Posted by bassman111

I've gotten some excellent help on these forums today. A HUGE thank you to everyone who assisted.

3 more questions giving me issues. If you could at least help get me started I would appreciate that...no need to do all my homework for me lol

Find the derivative of the following functions-

f(t)=tan^-1(e^4t-2)tan(e^t+1)

g(z)=ln(z^2)-5ze^(2z-1)

Given e^(x^2)+y^6=3xtan(y), find y'

I'm not really sure where to start on these

I can't remember the derivative of a inverse tangent, but just look it up and use the chain rule. For example, to find g'(z), use the chain rule:

You should know that the derivative of a function $f(u)=ln(u)$ is just $\frac{1}{u}$

So to differentiate g(z):

$g'(z)=\frac{2z}{z^2}-(5e^{2z-1}+10ze^{2z-1})$

$=\frac{2}{z}-e^{2z-1}(5+10z)$

Do you see how I applied the power rule and the chain rule?

**bassman111** · October 18th 2009, 04:24 PM

I understand the product and the power rule...the chain rule still confuses me a bit though.

**adkinsjr** · October 18th 2009, 04:30 PM

Originally Posted by bassman111

I understand the product and the power rule...the chain rule still confuses me a bit though.

Ok, if we have the function:

$f(z)=ln(z^2)$

This can be expressed as a composite function. $f(u(z))=ln(u)$ where $u=z^2$

$\frac{df}{dz}=\frac{df}{du}\frac{du}{dz}$

$\frac{df}{du}=\frac{1}{u}=\frac{1}{z^2}$

$\frac{du}{dz}=2z$

If we multiply the two derivatives above we have $\frac{2z}{z^2}=\frac{2}{z}$

This is how I used the chain rule to get the first term in: $=\frac{2}{z}-e^{2z-1}(5+10z)$

**bassman111** · October 18th 2009, 04:36 PM

Originally Posted by adkinsjr

I can't remember the derivative of a inverse tangent, but just look it up and use the chain rule. For example, to find g'(z), use the chain rule:

You should know that the derivative of a function $f(u)=ln(u)$ is just $\frac{1}{u}$

So to differentiate g(z):

$g'(z)=\frac{2z}{z^2}-(5e^{2z-1}+10ze^{2z-1})$

$=\frac{2}{z}-e^{2z-1}(5+10z)$

Do you see how I applied the power rule and the chain rule?

so is $=\frac{2}{z}-e^{2z-1}(5+10z)$ the derivative? or do I need to eliminate 'e' from the equation?

**adkinsjr** · October 18th 2009, 04:51 PM

Originally Posted by bassman111

so is $=\frac{2}{z}-e^{2z-1}(5+10z)$ the derivative? or do I need to eliminate 'e' from the equation?

Yes, that is the derivative. Whenever you differentiate the exponential function $e^u$ you will get $\frac{du}{dz}e^{u}$ where $u=u(z)$

**bassman111** · October 18th 2009, 04:57 PM

ok, so if x=e^4t-2 and y=e^t+1

I have...
(1/(1+x))*(sec^2y)
for the first question. Is this correct so far? And where would I go from here...?

**bassman111** · October 18th 2009, 05:34 PM

double post

**adkinsjr** · October 18th 2009, 05:49 PM

Originally Posted by bassman111

ok, so if x=e^4t-2 and y=e^t+1

I have...
(1/(1+x))*(sec^2y)
for the first question. Is this correct so far? And where would I go from here...?

You have to apply the power rule to differentiate $tan^{-1}(x)tan(y)$ , but then you also have to consider that x and y are both composite functions of t, which means you have apply the chain rule to both terms.

The derivative start as:

$\frac{dx}{dt}\frac{1}{1+x^2}tan(y)+\frac{dy}{dt}ta n(x)\frac{d}{dy}tan(y)$

**bassman111** · October 18th 2009, 05:58 PM

:\ either my teacher did not do a good job explaining or I'm not very good at this because what you responded with just does not make sense to me...sorry. Things are not looking good for this assignment

**adkinsjr** · October 18th 2009, 05:59 PM

Sorry, not "POWER RULE," I meant product rule.

**adkinsjr** · October 18th 2009, 06:11 PM

Originally Posted by bassman111

:\ either my teacher did not do a good job explaining or I'm not very good at this because what you responded with just does not make sense to me...sorry. Things are not looking good for this assignment

Calculus isn't easy. It took the genius of Isaac Newton to develop, so don't be discouraged.

Do you understand why you have to apply the product rule to $tan^{-1}(x)tan(y)$ ?

Remember that you defined x and y as functions of t when you wrote:

$x=e^{4t-2}$ and $y=e^{t+1}$

This means that $tan^{-1}(x)$ is a function of a function. The same is true with $tan(y)$ . In order to differentiate I have to apply the chain rule.

If we consider the function:

$tan^{-1}(x)$

we have the derivative:

$\frac{dx}{dt}\frac{1}{1+x^2}$

$\frac{dx}{dt}=4e^{4t-2}$

$4e^{4t-2}{1+x^2}$

You don't want leave it in this form since $x=e^{4t-2}$ , you should substitute this for the $x^2$

Can you tell what the derivative with respect to t of $tan(y)$ is?

**bassman111** · October 18th 2009, 06:13 PM

Originally Posted by adkinsjr

Calculus isn't easy. It took the genius of Isaac Newton to develop, so don't be discouraged.

Do you understand why you have to apply the power rule to $tan^{-1}(x)tan(y)$ ?

Remember that you defined x and y as functions of t when you wrote:

$x=e^{4t-2}$ and $y=e^{t+1}$

This means that $tan^{-1}(x)$ is a function of a function. The same is true with $tan(y)$ . In order to differentiate I have to apply the chain rule.

$tan^{-1}(x)$

Then the derivative is:

$\frac{dx}{dt}\frac{1}{1+x^2}$

$\frac{dx}{dt}=4e^{4t-2}$

$4e^{4t-2}{1+x^2}$

You don't want leave it in this form since $x=e^{4t-2}$ , you should substitute this for the $x^2$

Can you tell what the derivative with respect to t of $tan(y)$ is?

sec^2(e^t+1)
right?

**adkinsjr** · October 18th 2009, 06:28 PM

Originally Posted by bassman111

sec^2(e^t+1)
right?

Exactly. Now you have the derivatives of $tan^{-1}(x)$ and $tan(y)$

So the derivative of the product $tan^{-1}(x)tan(y)$ is given by the product rule. Do you see how to apply the product rule to this?

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