godorovg
3x^3+2x^2 +X+5 the derivative
evalauate y' when x=3 need a little help bit rusty
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godorovg
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in order to the first you lower the power by one like this
3x^2+2x^1+5 that is der right?
UnkleRhaukus
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you also have to divide by the original index
godorovg
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you divide by 3 because of 3x^3
cornitodisc
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x^n=nx^n-1
godorovg
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so if I divide by 3, than that would make this x^2 instead of 3x^2 I am on the right track?
UnkleRhaukus
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opps ive got mixed up
godorovg
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y' 3x^2 is X^2 right?
UnkleRhaukus
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i should have said times by the index
\[y(x)=ax^n\qquad\Rightarrow\qquad y'(x)=anx^{n-1}\]
UnkleRhaukus
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so sorry if i confused you
godorovg
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so times by index makes 3x^3 Y' 6x^2 right?
UnkleRhaukus
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not quite
\[y(x)=3x^3\qquad y'(x)=3\times3x^2\]
godorovg
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here is what I know y' when doing this you lower the power by one and something??
UnkleRhaukus
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when taking the derivative multiply by the power and then lower the power
godorovg
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so you are telling that power before lowering divide to y'
godorovg
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like this 4x^3 so that would be 'y 4 right?
godorovg
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4x
UnkleRhaukus
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i ment times when i said divide originally , dont divide when taking the derivative
godorovg
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can we start over plz..
UnkleRhaukus
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yes ,
godorovg
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y' 4x^2 would y'4x
UnkleRhaukus
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\[y(x)=ax^n\qquad\Rightarrow\qquad y'(x)=n\times ax^{n-1}\]
\[y(x)=4x^2\qquad\Rightarrow\qquad y'(x)=2\times 4x^{2-1}\]
godorovg
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y'(x) would be 8x
UnkleRhaukus
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thats right now
godorovg
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okay so the power is what you are times before lowering the power?
UnkleRhaukus
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that is right
godorovg
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so for another example 10X^5 y'(x) 50x^4
UnkleRhaukus
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you got it \(\checkmark\)
godorovg
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so back to the problem now
UnkleRhaukus
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\[y(x)=3x^3+2x^2 +x+5\]\[y'(x)=\]
godorovg
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y'(x) 6x+4X +5
UnkleRhaukus
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not quite,
note that
\[x=x^1\]\[5=5x^0\]
godorovg
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y'(x) 9x+4+5
UnkleRhaukus
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try again
godorovg
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y'(x) 6x^2 +4x+5 +1
UnkleRhaukus
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ok lets do this one term at a time
\[(3x^3)^\prime=\]
godorovg
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y'(x) 9x^2
UnkleRhaukus
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right,
now
\[(2x^2)^\prime=\]
godorovg
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y'(x) 9x^2 +2x
godorovg
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y'(x) 9x^2 4+5+1
UnkleRhaukus
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nope
godorovg
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y'(x) 9x^2 +4X+5
UnkleRhaukus
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you have the first two terms right,
UnkleRhaukus
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\[f(x)=x=x^1\]
\[f^\prime(x)=1\times x^{1-1}=x^0=1\]
godorovg
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so y'(x) 9x^2+4x+1+5
UnkleRhaukus
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the derivative of the last term
\[g(x)=5=5x^0\]
\[g^\prime(x)=0\times 5x^{0-1}=\]
godorovg
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is 0
UnkleRhaukus
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yeah the derivative of a constant is always zero
godorovg
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so the end number unless it has a power drop off?
UnkleRhaukus
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thats right
UnkleRhaukus
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\[f(x)=x^2+5\qquad f'(x)=2x\]\[g(x)=x^2+9\qquad g'(x)=2x\]\[h(x)=x^2−71\qquad f'(x)=2x\]
UnkleRhaukus
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that last one is ment to be h'
godorovg
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cool love it so far because in math you can't drop anything, in calc 1 you can this is way cool..
godorovg
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do you have time to do one more?
UnkleRhaukus
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yeah can you get the answer to you original question now?
UnkleRhaukus
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\[y(x)=3x^3+2x^2 +x+5\]
\[y~'(x)=\]
godorovg
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y'(x) 9x^2+4x+1
UnkleRhaukus
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\[\Large\color\red\checkmark\]
godorovg
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okay any other tips here?
UnkleRhaukus
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\[y'(x)=\frac{\text dy(x)}{\text dx}\]
godorovg
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explain a little??
UnkleRhaukus
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its just a different notation, they men the same thing
UnkleRhaukus
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the second has a bit more information because it explicitly say that we are to take the derivative of the function with respect to x, ie we look at the power of the x terms,
godorovg
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dx(x)^1?dx
godorovg
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so the power on the is 1 right?
godorovg
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just to let know, I am in college not high school
UnkleRhaukus
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\[y(x)=x=x^1\]
\[\frac{\text dy(x)}{\text dx}=y'(x)=1\times x^{1-1}=x^0=1\]
UnkleRhaukus
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im not sure what you mean by this
dx(x)^1?dx
godorovg
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I meant to use / not?
UnkleRhaukus
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dx(x)^1/dx \[=\large \frac{\text d(x)^1}{\text dx}\]?
godorovg
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yes that is what I meant to type
UnkleRhaukus
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\[\frac{\text d(x)}{\text dx}=\frac{\text d(x^1)}{\text dx}=1\times x^0=1\]
UnkleRhaukus
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im not sure if i am answer your question?
godorovg
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okay now for the last question here how do this with y"(x) sin
godorovg
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yes you have
UnkleRhaukus
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should we evaluate y' in the first question when x=3 before that?
godorovg
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y'(x) 9x^2+4x+1 so plug in 3 so we have y'(x) 9x^2 (3) +4x(3) +1 which is y'(x) 27x^2+ 12X +1
UnkleRhaukus
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\[y(x)=3x^3+2x^2+x+5\]
\[y'(x)=9x^2+4x+1\]
\[y'(3)=9(3)^2+4(3)+1=9\times3\times3+4\times3+1\]
godorovg
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y'(3) 81x^2+12+1
godorovg
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12x
UnkleRhaukus
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when you evaluate a function at a particular x value you substitute this value for x in the equation, there should be no x in left when evaluated
godorovg
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sorry i mess me up by not putting the like this (3)^2
godorovg
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so the answer is F'(3) 81x^2 + 12X +1