anonymous
  • anonymous
derivative of y=x^(3/2)log(base3)squareroot(x+7)
Calculus1
chestercat
  • chestercat
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zepdrix
  • zepdrix
\(\large\rm y=x^{3/2}\cdot\log_3\sqrt{x+7}\) This guy? Hmm this looks like a fun one :) lol
anonymous
  • anonymous
\[y= x ^{3/2}\log_{3}\sqrt{x+7}\]
anonymous
  • anonymous
yeah lol

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zepdrix
  • zepdrix
Do you remember your change of base formula for logs?\[\large\rm \log_{a}(b)=\frac{\ln(b)}{\ln(a)}\]We would choose natural log like this ^ Because it's easier to differentiate.
zepdrix
  • zepdrix
Unless you have a special formula memorized for log derivatives :) \[\large\rm y=x^{3/2}\cdot\frac{\ln\sqrt{x+7}}{\ln3}\]
anonymous
  • anonymous
so it would be \[(3/2)x ^{1/2}*(\ln \sqrt{x+7}/\ln3)\]
anonymous
  • anonymous
?
zepdrix
  • zepdrix
No no no :O Product rule.
anonymous
  • anonymous
what should i do then?
zepdrix
  • zepdrix
I'll do the "set up" and you can let me know if it makes sense:\[\large\rm y'=\color{royalblue}{\left(x^{3/2}\right)'}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}\]This is our product rule "set up". I haven't taken any derivatives yet. I'm just setting it up. We need to differentiate the blue parts.
anonymous
  • anonymous
is the first blue part \[(3/2)x ^{(3/2)-1}\]
zepdrix
  • zepdrix
\[\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}\]Ok ya that sounds right!
zepdrix
  • zepdrix
This other guy is a bit tricky though, ya? :)
anonymous
  • anonymous
\[1/((1/2)x ^{-1/2}+(1/2)7^{-1/2}) /(1/3)\]
anonymous
  • anonymous
?
zepdrix
  • zepdrix
No. And oh boy, did you ever make things complicated :D lol
zepdrix
  • zepdrix
Let's back up a sec, apply a helpful log rule before we take the derivative.
anonymous
  • anonymous
sorry lol
anonymous
  • anonymous
\[\ln \sqrt{x+7}-\ln 3\]
anonymous
  • anonymous
?
zepdrix
  • zepdrix
First notice that ln3 is just a constant, so we can pull it out of the differentiation process and ignore it.\[\large\rm \color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}=\frac{1}{\ln3}\color{royalblue}{\left(\ln\sqrt{x+7}\right)'}\]
anonymous
  • anonymous
what about the x+7
zepdrix
  • zepdrix
Next rewrite your square root as a 1/2 power,\[\large\rm \frac{1}{\ln3}\color{royalblue}{\left(\ln(x+7)^{1/2}\right)'}\]And apply another log rule,\[\large\rm =\frac{1}{\ln3}\color{royalblue}{\left(\frac{1}{2}\ln(x+7)\right)'}\]And again pull the 1/2 out front because it's just a constant,\[\large\rm =\frac{1}{2\ln3}\color{royalblue}{\left(\ln(x+7)\right)'}\]The problem becomes a lot easier to deal with at this point.
zepdrix
  • zepdrix
This is one of the things that can make Calc a difficult subject, remember all of these silly Algebra rules :) They're really helpful though.
anonymous
  • anonymous
does the x+7 become 8
anonymous
  • anonymous
or does it become 2
zepdrix
  • zepdrix
I don't understand what you're asking
anonymous
  • anonymous
does \[(\ln(x+7))\prime = \ln(2)\prime = 1/2\]
zepdrix
  • zepdrix
?? 0_o
zepdrix
  • zepdrix
I have no idea where that 2 is coming from.. this is so confusing
zepdrix
  • zepdrix
Do you remember your derivative for ln(x) ?
anonymous
  • anonymous
isnt the derivative of a constant = 1?
anonymous
  • anonymous
so the derivative of x = 1 and the derivative of 7 = 1
anonymous
  • anonymous
so 1+1 = 2
anonymous
  • anonymous
???
zepdrix
  • zepdrix
The derivative of a constant is = 0. x is not a constant. But let's not worry about that. That's not what we're dealing with here. I'm asking if you know the derivative of \(\large\rm \ln(x)\)
anonymous
  • anonymous
is it 1?
zepdrix
  • zepdrix
... No :(
anonymous
  • anonymous
isnt the derivative of ln(x) = 1/x
zepdrix
  • zepdrix
Yes.
zepdrix
  • zepdrix
So then by that same rule,\[\large\rm \frac{d}{dx}\ln(x+7)=\frac{1}{(x+7)}\]
anonymous
  • anonymous
wait so is the answer \[(3/2)x ^{1/2}-((\ln(\sqrt{x+7})/\ln3)*x ^{3/2})*(1/x+7)\]
zepdrix
  • zepdrix
No :(
anonymous
  • anonymous
what is the right answer?
zepdrix
  • zepdrix
By the steps I was showing you before, the derivative of that second blue thing we were working on, would be,\[\large\rm \color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}=\color{orangered}{\frac{1}{2\ln3(x+7)}}\]
zepdrix
  • zepdrix
\[\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}\] So I guess our final result is something like this,\[\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{orangered}{\frac{1}{(2\ln3)(x+7)}}\]
zepdrix
  • zepdrix
Practice more problems cp :O Really really need to study!
anonymous
  • anonymous
OH OKAY
anonymous
  • anonymous
I GET IT NOW
anonymous
  • anonymous
THANK YOU SO MUCH

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