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3x^3+2x^2 +X+5 the derivative evalauate y' when x=3 need a little help bit rusty

Mathematics
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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?
you also have to divide by the original index
you divide by 3 because of 3x^3

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Other answers:

x^n=nx^n-1
so if I divide by 3, than that would make this x^2 instead of 3x^2 I am on the right track?
opps ive got mixed up
y' 3x^2 is X^2 right?
i should have said times by the index \[y(x)=ax^n\qquad\Rightarrow\qquad y'(x)=anx^{n-1}\]
so sorry if i confused you
so times by index makes 3x^3 Y' 6x^2 right?
not quite \[y(x)=3x^3\qquad y'(x)=3\times3x^2\]
here is what I know y' when doing this you lower the power by one and something??
when taking the derivative multiply by the power and then lower the power
so you are telling that power before lowering divide to y'
like this 4x^3 so that would be 'y 4 right?
4x
i ment times when i said divide originally , dont divide when taking the derivative
can we start over plz..
yes ,
y' 4x^2 would y'4x
\[y(x)=ax^n\qquad\Rightarrow\qquad y'(x)=n\times ax^{n-1}\] \[y(x)=4x^2\qquad\Rightarrow\qquad y'(x)=2\times 4x^{2-1}\]
y'(x) would be 8x
thats right now
okay so the power is what you are times before lowering the power?
that is right
so for another example 10X^5 y'(x) 50x^4
you got it \(\checkmark\)
so back to the problem now
\[y(x)=3x^3+2x^2 +x+5\]\[y'(x)=\]
y'(x) 6x+4X +5
not quite, note that \[x=x^1\]\[5=5x^0\]
y'(x) 9x+4+5
try again
y'(x) 6x^2 +4x+5 +1
ok lets do this one term at a time \[(3x^3)^\prime=\]
y'(x) 9x^2
right, now \[(2x^2)^\prime=\]
y'(x) 9x^2 +2x
y'(x) 9x^2 4+5+1
nope
y'(x) 9x^2 +4X+5
you have the first two terms right,
\[f(x)=x=x^1\] \[f^\prime(x)=1\times x^{1-1}=x^0=1\]
so y'(x) 9x^2+4x+1+5
the derivative of the last term \[g(x)=5=5x^0\] \[g^\prime(x)=0\times 5x^{0-1}=\]
is 0
yeah the derivative of a constant is always zero
so the end number unless it has a power drop off?
thats right
\[f(x)=x^2+5\qquad f'(x)=2x\]\[g(x)=x^2+9\qquad g'(x)=2x\]\[h(x)=x^2−71\qquad f'(x)=2x\]
that last one is ment to be h'
cool love it so far because in math you can't drop anything, in calc 1 you can this is way cool..
do you have time to do one more?
yeah can you get the answer to you original question now?
\[y(x)=3x^3+2x^2 +x+5\] \[y~'(x)=\]
y'(x) 9x^2+4x+1
\[\Large\color\red\checkmark\]
okay any other tips here?
\[y'(x)=\frac{\text dy(x)}{\text dx}\]
explain a little??
its just a different notation, they men the same thing
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,
dx(x)^1?dx
so the power on the is 1 right?
just to let know, I am in college not high school
\[y(x)=x=x^1\] \[\frac{\text dy(x)}{\text dx}=y'(x)=1\times x^{1-1}=x^0=1\]
im not sure what you mean by this dx(x)^1?dx
I meant to use / not?
dx(x)^1/dx \[=\large \frac{\text d(x)^1}{\text dx}\]?
yes that is what I meant to type
\[\frac{\text d(x)}{\text dx}=\frac{\text d(x^1)}{\text dx}=1\times x^0=1\]
im not sure if i am answer your question?
okay now for the last question here how do this with y"(x) sin
yes you have
should we evaluate y' in the first question when x=3 before that?
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
\[y(x)=3x^3+2x^2+x+5\] \[y'(x)=9x^2+4x+1\] \[y'(3)=9(3)^2+4(3)+1=9\times3\times3+4\times3+1\]
y'(3) 81x^2+12+1
12x
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
sorry i mess me up by not putting the like this (3)^2
so the answer is F'(3) 81x^2 + 12X +1

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