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

- anonymous

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

- chestercat

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- zepdrix

\(\large\rm y=x^{3/2}\cdot\log_3\sqrt{x+7}\)
This guy?
Hmm this looks like a fun one :) lol

- anonymous

\[y= x ^{3/2}\log_{3}\sqrt{x+7}\]

- anonymous

yeah lol

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## More answers

- 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

Unless you have a special formula memorized for log derivatives :)
\[\large\rm y=x^{3/2}\cdot\frac{\ln\sqrt{x+7}}{\ln3}\]

- anonymous

so it would be \[(3/2)x ^{1/2}*(\ln \sqrt{x+7}/\ln3)\]

- anonymous

?

- zepdrix

No no no :O
Product rule.

- anonymous

what should i do then?

- 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

is the first blue part \[(3/2)x ^{(3/2)-1}\]

- 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

This other guy is a bit tricky though, ya? :)

- anonymous

\[1/((1/2)x ^{-1/2}+(1/2)7^{-1/2}) /(1/3)\]

- anonymous

?

- zepdrix

No.
And oh boy, did you ever make things complicated :D lol

- zepdrix

Let's back up a sec, apply a helpful log rule before we take the derivative.

- anonymous

sorry lol

- anonymous

\[\ln \sqrt{x+7}-\ln 3\]

- anonymous

?

- 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

what about the x+7

- 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

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

does the x+7 become 8

- anonymous

or does it become 2

- zepdrix

I don't understand what you're asking

- anonymous

does \[(\ln(x+7))\prime = \ln(2)\prime = 1/2\]

- zepdrix

?? 0_o

- zepdrix

I have no idea where that 2 is coming from.. this is so confusing

- zepdrix

Do you remember your derivative for ln(x) ?

- anonymous

isnt the derivative of a constant = 1?

- anonymous

so the derivative of x = 1 and the derivative of 7 = 1

- anonymous

so 1+1 = 2

- anonymous

???

- 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

is it 1?

- zepdrix

... No :(

- anonymous

isnt the derivative of ln(x) = 1/x

- zepdrix

Yes.

- zepdrix

So then by that same rule,\[\large\rm \frac{d}{dx}\ln(x+7)=\frac{1}{(x+7)}\]

- anonymous

wait so is the answer \[(3/2)x ^{1/2}-((\ln(\sqrt{x+7})/\ln3)*x ^{3/2})*(1/x+7)\]

- zepdrix

No :(

- anonymous

what is the right answer?

- 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

\[\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}\]
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

Practice more problems cp :O
Really really need to study!

- anonymous

OH OKAY

- anonymous

I GET IT NOW

- anonymous

THANK YOU SO MUCH

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