Question

How to use leibniz notation to solve

y=sqr 2u , u=6-x x=-3

Answer 1

Wait is the sqr representing a square-root symbol?

Answer 2

yea

Answer 3

i did the question. I got (u^-1/2)(2)(-1)

Answer 4

So it's:

\[y=\sqrt{2u}, u=6-x,x=-3\]Right? @0213

Answer 5

yes

Answer 6

The result would be a constant then, so the answer would be 0, correct?

Answer 7

the answer at the back of my book is 1/2sqr3

Answer 8

sorry its -1/3sqr2

Answer 9

So first, replace u with 6-x in the square root:\[y=\sqrt{2(6-x)} \rightarrow y=\sqrt{12-2x}\]Now just differentiate with power rule and chain rule.\[y'=\frac{ 1 }{ 2 }(12-2x)^{-\frac{ 1 }{ 2 }}*-2 \rightarrow y'=-\frac{ 2 }{ 2\sqrt{12-2x} } \rightarrow y'=-\frac{ 1 }{ \sqrt{12-2x} }\]

Answer 10

And the derivative here at x = -3 is then:\[y'(-3)=-\frac{ 1 }{ \sqrt{12-2(-3)} }\rightarrow y'(-3)=-\frac{ 1 }{ \sqrt{12+6} } \rightarrow y'(3)=-\frac{ 1 }{ \sqrt{18} }\]

Answer 11

@0213

Answer 12

but i have to use leibniz notation. I have to show the derivative of y multplied by the derivative of u . Thats the way I'm doing it .

Answer 13

Oh you want to show it in terms of du/dx?

Answer 14

yes

Answer 15

Should've said that before -.-

\[y'=\frac{ 1 }{ 2 }(2u)^{-\frac{ 1 }{ 2 }}*2*\frac{ du }{ dx } \rightarrow y'=\frac{ 1 }{ \sqrt{2u} }*\frac{ du }{ dx } \rightarrow y'=\frac{ 1 }{ \sqrt{12-2x} }*\frac{ du }{ dx }\]@0213

Answer 16

so I won't multiply the 2 and the 1/2 when I bring it down in the first part of the equation

Answer 17

First part is power rule and chain rule. You take the power of 1/2 to the front of 2u and the power of 2u becomes -1/2, that's the power rule. Now you have to differentiate the stuff under the square root as well and multiply it by that. This is the chain rule. So when you differentiate 2u with constant multiple rule, you get 2*du/dx. So now you multiply the whole thing by 2*du/dx but the 2 in the denominator of 1/2 and the 2 being multiplied by du/dx cancel out. From there you just simplify and in the last step, I just replaced the u in the square root with 6 - x to make it the root of 12 - 2x. @0213

Answer 18

so when I have a square root with a value like 3x or 5x-1 , I have to look at the square root alone and take the derivative of it and then look at the value inside and take the derivative of it