derivative of x(3x-16)/(2 root(x-4)^3)
\[y =\frac{ x(3x -16) }{ 2\sqrt{(x -4)^{3}} }\]Is this what the equation is?
yes.
\[\frac{ dy }{ dx }=\frac{ (6x -16)(2\sqrt{(x -4)^{3}})-3x(3x -16)\sqrt{x -4} }{ 4(x -4)^{3} }\]Do you understand this first step or would you like me to break it down further?
I understand this first step.
Now do you see the common factor of \[\sqrt{x -4}\]in the numerator?
yes
OK so then show me the next step!
are you asking me what the common facotr of root x-4 is or asking me if I see root x-4?
\[\frac{ dy }{ dx }=\frac{ \sqrt{x -4}[2(6x -16)(x -4)-3x(3x -16)] }{ 4(x -4)^{3} }\]Do you understand what I meant, now?
oh yes I do.
now what?
Do you know how to simplify from there?
would I first distribute what I can in the numerator first?
Yes, distribute inside the brackets and simplify. The root of (x - 4) Which is (x - 4)^(1/2) can be divided by the (x - 4) in the denominator.
what do you mean the root of (x-4) can be divided by the x-4 in the denominator?
would the inside brackets be 3x^2-80x+80
Sorry, I was busy with something.
no worries. be right back, I have to eat dinner.
No that is incorrect.
That's OK. I have to log out so I'll post the final steps for you before I go.
\[\frac{ dy }{ dx }=\frac{ 2(6x ^{2}-40x +64)-9x ^{2} +48x}{ 4\sqrt{(x -4)^{5}} }\] \[\frac{ dy }{ dx }=\frac{ 12x ^{2}-80x +128-9x ^{2}+48x }{ 4(\sqrt{(x -4)^{5}} }\] \[\frac{ dy }{ dx }=\frac{ 3x ^{2}-32x +128 }{ 4\sqrt{(x -4)^{5}} }\]
so how did you get rid of the root x-4 to get 4(root x-4)^5? I got the numerator after redo-ing everything.
@xkat If you divide\[\frac{ (x -4)^{\frac{ 1 }{ 2 }} }{ (x -4)^{3} }\]What do you get? well, whenever you divide two powers with the same base, you subtract their exponents. Correct? The exponents are 1/2 and 3, so when you subtract these exponents, you get an exponent that is -5/2, correct? Now express the final answer with positive exponents, thus that is how end with\[\sqrt{(x -4)^{5}}\]in the denominator.
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