Logarithmic Differentiation Check

[Silvanoshei]

Find dy/dx:

$\displaystyle y=x^{x+1}$

My work: Is it correct?
Edit: I forgot product rule didn't I?

$\displaystyle lny=lnx^{x+1}$
$\displaystyle lny=(x+1)lnx$
$\displaystyle \frac{d}{dx}lny=(x+1)\frac{d}{dx}lnx$
$\displaystyle \frac{1}{y}\frac{dy}{dx}=(x+1)(\frac{1}{x}*1)$
$\displaystyle \frac{dy}{dx}=(x+1)(\frac{1}{x})(x^{x+1})$


Corrected Answer?
$\displaystyle \frac{1}{y}\frac{dy}{dx}=(x+1)\frac{d}{dx}(lnx)+(lnx)\frac{d}{dx}(x+1)$
$\displaystyle =(x+1)(\frac{1}{x})+(lnx)$
$\displaystyle \frac{dy}{dx}=(\frac{x+1}{x})+(lnx)(x^{x+1})$

 

[Deleted member 4993]

Find dy/dx:

$\displaystyle y=x^{x+1}$

My work: Is it correct?

$\displaystyle lny=lnx^{x+1}$
$\displaystyle lny=(x+1)lnx$
$\displaystyle \frac{d}{dx}lny=(x+1)\frac{d}{dx}lnx$
$\displaystyle \frac{1}{y}\frac{dy}{dx}=(x+1)(\frac{1}{x}*1)$
$\displaystyle \frac{dy}{dx}=(x+1)(\frac{1}{x})(x^{x+1})$

Correct ... excellent work...

No ... I am wrong... see below ... going to the corner

Hope Jeff is still there ... otherwise it is going to be lonely...

 

[lookagain]

Correct ... excellent work...

Silvanoshei and Subhotosh Khan, that is not correct. Beginning at the third line of work, the solution is wrong. That is not the correct derivative on the right side of the equation. The product rule is the expected method at that point. Please apply the product rule to the expression (x + 1)*ln(x) for that step. As a consequence, the final answer is also incorrect.

 

[srmichael]

Find dy/dx:

$\displaystyle y=x^{x+1}$

My work: Is it correct?

$\displaystyle ln(y)=lnx^{x+1}$
$\displaystyle lny=(x+1)lnx$
$\displaystyle \frac{d}{dx}lny=(x+1)\frac{d}{dx}lnx$ ===> Not Quite (use product rule)
$\displaystyle \frac{1}{y}\frac{dy}{dx}=(x+1)(\frac{1}{x}*1)$
$\displaystyle \frac{dy}{dx}=(x+1)(\frac{1}{x})(x^{x+1})$

$\displaystyle y=x^{x+1}$
$\displaystyle ln(y)=ln(x)^{x+1}$
$\displaystyle ln(y)=(x+1)ln(x)$
$\displaystyle \frac{d}{dx}ln(y)=(x+1)\frac{d}{dx}ln(x)+ln(x) \frac{d}{dx}(x+1)$
$\displaystyle \frac{1}{y}\frac{dy}{dx}=(x+1)(\frac{1}{x}*1)+ln(x)*(1)$
$\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{(x+1)}{x}+ln(x)$
$\displaystyle \frac{dy}{dx}=y[\frac{(x+1)}{x}+ln(x)]$
$\displaystyle \frac{dy}{dx}=x^{x+1}[1+\frac{1}{x}+ln(x)]$

 

[Silvanoshei]

Darn it, I knew it lol. Thanks guys.

 

[Deleted member 4993]

Find dy/dx:

$\displaystyle y=x^{x+1}$

My work: Is it correct?
Edit: I forgot product rule didn't I?

$\displaystyle lny=lnx^{x+1}$
$\displaystyle lny=(x+1)lnx$
$\displaystyle \frac{d}{dx}lny=(x+1)\frac{d}{dx}lnx$
$\displaystyle \frac{1}{y}\frac{dy}{dx}=(x+1)(\frac{1}{x}*1)$
$\displaystyle \frac{dy}{dx}=(x+1)(\frac{1}{x})(x^{x+1})$


Corrected Answer? Nope.. needs [] brackets as shown below
$\displaystyle \frac{1}{y}\frac{dy}{dx}=(x+1)\frac{d}{dx}(lnx)+(lnx)\frac{d}{dx}(x+1)$
$\displaystyle =(x+1)(\frac{1}{x})+(lnx)$
$\displaystyle \frac{dy}{dx}=[(\frac{x+1}{x})+(lnx)](x^{x+1})$

.

 

[Silvanoshei]

Problem #2

$\displaystyle y=(sinx)^{6x}$

My work: Edited corrections in red.

$\displaystyle lny=ln(sinx)^{6x}$
$\displaystyle \frac{1}{y}\frac{d}{dx}=6xln(sinx)$
$\displaystyle \frac{1}{y}\frac{d}{dx}=6x\frac{d}{dx}ln(sinx)+ln(sinx)\frac{d}{dx}(6x)$
$\displaystyle \frac{1}{y}\frac{d}{dx}=6x(\frac{1}{sinx})(cosx)+ln(sinx)(6)$
$\displaystyle \frac{dy}{dx}=sinx^{6x}(\frac{6xcosx}{sinx}+6ln(sinx))$

 

[srmichael]

Problem #2

$\displaystyle y=(sinx)^{6x}$

My work: Edited corrections in red.

$\displaystyle lny=ln(sinx)^{6x}$
$\displaystyle \frac{1}{y}\frac{d}{dx}=6xln(sinx)$
$\displaystyle \frac{1}{y}\frac{d}{dx}=6x\frac{d}{dx}ln(sinx)+ln(sinx)\frac{d}{dx}(6x)$
$\displaystyle \frac{1}{y}\frac{d}{dx}=6x(\frac{1}{sinx})(cosx)+ln(sinx)(6)$
$\displaystyle \frac{dy}{dx}=sinx^{6x}(\frac{6xcosx}{sinx}+6ln(sinx))$

FIRST: One problem per thread!

That being said, this looks good! You can do one last step and convert cosx/sinx to cotx. Keep this in mind had this been a multliple choice problem on a test. Also, make sure there are parentheses around sinx or else it looks like the angle x is being raised to the 6x power and not sin(x). The answer would have probably been:

$\displaystyle \frac{dy}{dx}=(sinx)^{6x}[6xcotx+6ln(sinx)]$

or possibly

$\displaystyle \frac{dy}{dx}=6(sinx)^{6x}[xcotx+ln(sinx)]$

 

[Silvanoshei]

Thanks srmichael! Sorry about that, I looked at the rules quick on posting and didn't see that catch my eye about 1 problem per thread. Noted and thanks for the help!

 

[srmichael]

Thanks srmichael! Sorry about that, I looked at the rules quick on posting and didn't see that catch my eye about 1 problem per thread. Noted and thanks for the help!

No sweat. But you know what, I just checked the guidelines and I, too, do not see any mentioning of one question per thread unless I just flat out missed it. Interesting.