Question

find dy/dx for y=1+3x-x2

Answer 1

3-2x

Answer 2

can you please explain how you have done this

Answer 3

The power rule states that the derivative of an exponential, such as 2x^2 is the coefficient times the exponent, whilst the exponent is subtracted by one. So, in my exampled you'd get 4x. Constants equal to zeros. As in, 1+2x^2 would just be 4x. Apply that to your problem.

Answer 4

i got your point...
see i have also did same but my answer is coming 3-2=1 not 3-2x

Answer 5

can you please elaborate it

Answer 6

That last value in your poblem, is it x2 or x^2?

Answer 7

x square

Answer 8

So, the 1 becomes a zero. the 3x loses the x, and just imagine the x^2 looked like this: -1x^2, multiply 2 by the -1, and then subtract the 2 by 1. You get -2x.

Answer 9

i am sending you the way i have done the problem please check it

Answer 10

Dude, forget the dx and dys. You're complicating yourself. Just multiply the exponents by their coefficients and subtract said exponents by one. That's it. x2 should be written x^2.

Answer 11

5x^5-3x^3+2x+100 = 25x^4-9x^2+2 <--- see it?

Answer 12

\(1+3x-x^2=18x^0+3x^1-x^2\)
so
\(\frac{d}{dx}(1+3x-x^2)=0*1x^{0-1}+1*3x^{1-1}-2x^{2-1}=0+3x^0-2x^1=3-2x\)

Answer 13

that shuold say 1 not 18...

Answer 14

@zz0c3k you have not used dy and dx???/

Answer 15

dx and dy just refer to the derivative in respect to those. Unless you get into implicit differentiation, they are kind of useless when doing something like what you are doing.

Answer 16

the question you gave me above i tried to solve it and got the ans 25x\[25x ^{4} - 9x ^{2} +2x+100\]

Answer 17

Numbers without letters = 0

Answer 18

Numbers with only one letters are just those numbers. 2x = 2.

Answer 19

\(\frac{dy}{dx}\) is the same thing as \(f(x)\) is a sense

if y = f(x) we have
\(\frac{d(f(x))}{dx}\)
so when we take the derivative we note this by \(\frac{d}{dx}(the \ function)=\frac{d}{dx}(f(x))\)

example

let \(y = f(x) = x^2\)
then we write
\(\frac{dy}{dx}=\frac{d}{dx}(x^2)=2x\)

Answer 20

but my professor told that we cannot ignore x we have to take 2x

Answer 21

So see it like this 2x^1. That would be 2(1)x^1-1 = 2x^0 and x^0 =1 so 2(1) = 2

Answer 22

see i got your point can you please solve another question

Answer 23

Why not try yourself?

Answer 24

a much better idea\(\huge \uparrow\)

Answer 25

i have tried bt i am getting same problem in the question the question says..
in each of the following case evaluate dy/dx when x=2
\[y=ax ^{2} +bx+c\]

Answer 26

@zzr0ck3r Jfc good luck.

Answer 27

please i need your help

Answer 28

if x = 2, then you are asking to take the derivative of a constant, what is the derivative of a constant?

Answer 29

i have typed what the question says we have to find dy/dx for the following equation and after finding we have to put the value of x

Answer 30

ahh, well
\(y'=2ax+b\)
so
\(4a+b\)

Answer 31

yaaa its correct how can you do it so fast!!!
please elaborate it

Answer 32

practice and it becomes second nature

Answer 33

ax^2+bx+c
we move the 2 down and decrement it by 1 and get 2ax^1 = 2ax
then we look at the bx term, the derivative is just the constant, so we have
b
so
2ax+b

Answer 34

the c term goes to 0

Answer 35

the derivative b is constant how?

Answer 36

2ax.. but hows it possible as i think that dx and x gets cut and we are left with 2a only..please correct my point

Answer 37

the derivative of bx is b for all b....

Answer 38

I dont know what you mean by x and dx get cut, I think you are getting confused with notation

don't even think of the notation as dy/dx right now, just think of the derivative of f(x) is f'(x)

Answer 39

or explain your question more

Answer 40

see my solution