WTF am i doing wrong

derivative of sqrt(x) is -1/2*(x^-3/2)

>>8138947
sorry, i see what you're trying to do.

you simplified the equation wrong.

thank-you very kind sir

DELETE THIS

>>8138938
instead of trying to remember two different rules for the chain rule, just always use the multiplication rule.

uv' + u'v

since you have negative powers, it will always end up equivalent to the quotient chain rule, but it's less steps to go through, less chances to fuck up

>>8138938
life :^)

>>8138964
This.

My Laplace transform table has 13 transforms but I only memorized 7 and would do it again.

>>8138938

[math]
\displaystyle\frac{d}{dx}[\frac{sin(x)}{\sqrt(x)}] = \frac{d}{dx}[sin(x)*x^{-\frac{1}{2}}]
\\
\displaystyle\frac{d}{dx}[u*v] = \frac{d}{du}u*v + \frac{d}{dv}v*u
[/math]]

smudge yourself out of existence. NOW

>>8138938
This may help u in the future:

http://www.derivative-calculator.net/
http://www.integral-calculator.com/

>>8139842
Not the product rule. You don't differentiate u wrt u, likewise with v. You differentiate wrt x. I.e.
[math]\frac{d}{dx}[u][/math]
And it's less ambiguous to write it as
[math]\frac{du}{dx}[/math]

>>8139918

But anon:

[math]\displaystyle \frac{d}{dx}u = \frac{d}{dx}*u = \frac{du}{dx}[/math]

>>8139918

Also, it's better to write it this way:

[math]\displaystyle \frac{d}{dx}[u(x)*v(x)] = \frac{d}{dx}u(x)*v(x) + \frac{d}{dx}v(x)*u(x)[/math]

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>>8139925
But anon:

>>8139931

[math]\displaystyle \displaystyle\frac{d}{dx}[u*v] = \frac{d}{d\textbf{u}}\mathbf{u}*v + \frac{d}{d\textbf{v}}\textbf{v}*u = \frac{d}{d}*v + \frac{d}{d}*u = v + u[/math]


Geeze anon why am I doing all the work for you?

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8KB, 250x250px

>>8139936
Guys, he did it! He figured it out! Maths is solved, you can all go home now!

[math]\frac{d}{dx}[u*v]=\frac{du}{dx}*v+\frac{dv}{dx}*u[/math]

Or, if you prefer:
[math]y=uv[/math]
[math]y'=u'v+v'u[/math]

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2MB, 1280x720px

>>8139943

Anon-くん, I prefer this >>8139927

>>8139947
Too many brackets for my liking but whatever floats your boat fampai. And the lack of square brackets to specify what's being differentiated when it's in that form makes it ambiguous. Specifically, this kind of ambiguity.
[math]\frac{d}{dx}u(x)*v(x)=\frac{d}{dx}[u(x)*v(x)][/math]
[math]\frac{d}{dx}u(x)*v(x)=\frac{d}{dx}[u(x)]*v(x)[/math]

>>8139950

I'm j/k anon I love you and your Leibniz notation.