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anonymous
 one year ago
derivative of y=x^(3/2)log(base3)squareroot(x+7)
anonymous
 one year ago
derivative of y=x^(3/2)log(base3)squareroot(x+7)

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\rm y=x^{3/2}\cdot\log_3\sqrt{x+7}\) This guy? Hmm this looks like a fun one :) lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y= x ^{3/2}\log_{3}\sqrt{x+7}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Do 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
 one year ago
Best ResponseYou've already chosen the best response.2Unless you have a special formula memorized for log derivatives :) \[\large\rm y=x^{3/2}\cdot\frac{\ln\sqrt{x+7}}{\ln3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it would be \[(3/2)x ^{1/2}*(\ln \sqrt{x+7}/\ln3)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2No no no :O Product rule.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what should i do then?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I'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
 one year ago
Best ResponseYou've already chosen the best response.0is the first blue part \[(3/2)x ^{(3/2)1}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}\]Ok ya that sounds right!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2This other guy is a bit tricky though, ya? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[1/((1/2)x ^{1/2}+(1/2)7^{1/2}) /(1/3)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2No. And oh boy, did you ever make things complicated :D lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Let's back up a sec, apply a helpful log rule before we take the derivative.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln \sqrt{x+7}\ln 3\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2First notice that ln3 is just a constant, so we can pull it out of the differentiation process and ignore it.\[\large\rm \color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}=\frac{1}{\ln3}\color{royalblue}{\left(\ln\sqrt{x+7}\right)'}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Next 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
 one year ago
Best ResponseYou've already chosen the best response.2This 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
 one year ago
Best ResponseYou've already chosen the best response.0does the x+7 become 8

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I don't understand what you're asking

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does \[(\ln(x+7))\prime = \ln(2)\prime = 1/2\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I have no idea where that 2 is coming from.. this is so confusing

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Do you remember your derivative for ln(x) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0isnt the derivative of a constant = 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the derivative of x = 1 and the derivative of 7 = 1

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.0isnt the derivative of ln(x) = 1/x

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So then by that same rule,\[\large\rm \frac{d}{dx}\ln(x+7)=\frac{1}{(x+7)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait so is the answer \[(3/2)x ^{1/2}((\ln(\sqrt{x+7})/\ln3)*x ^{3/2})*(1/x+7)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is the right answer?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2By 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}\] 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
 one year ago
Best ResponseYou've already chosen the best response.2Practice more problems cp :O Really really need to study!
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