anonymous
  • anonymous
3x^3+2x^2 +X+5 the derivative evalauate y' when x=3 need a little help bit rusty
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
in order to the first you lower the power by one like this 3x^2+2x^1+5 that is der right?
UnkleRhaukus
  • UnkleRhaukus
you also have to divide by the original index
anonymous
  • anonymous
you divide by 3 because of 3x^3

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

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