Basic Derivatives Example 1


Uploaded by TheIntegralCALC on 12.02.2010

Transcript:
Hey guys!
Okay. This derivative problem is going to be slightly different from the other two.
The other two were written as f(x) equals some function.
This one is going to be
y = 2x^2 + 3x - 17.
And you might think, because this is y instead of f(x),
that it's different but it's actually not different at all.
Before, we said f(x) when we changed it to the derivative,the notation was f'(x).
So same thing here, instead of y, when we take the derivative,
it's going to be y' is the derivative.
So that's the notation. And then, we're going to go ahead and
find the derivative of this function one term at a time.
So first, we're going to deal with the 2x^2.
And the way that we do that is the coefficient, 2,
times the exponent, 2,
keep the x, and subtract 1 from the exponent,
so, 2 - 1 is 1, so, we can put that here.
And we'll go...we'll just go through and then we'll simplify.
So then, 3x, as we learned before, the derivative of 3x is simply 3.
But if you look at the... at the 1, which is the implied exponent there,
you can say plus 3, the coefficient,
times 1, the exponent,
leave the x, and then
1 - 1, we always subtract 1 from the exponent,
1 - 1 of course is 0,
so we can do that there. And then the derivative of
17 is 0,
because 17 is a constant, so that's going to go away.
And then we'll simplify this. So 2 times 2, 4,
and x, the 1 exponent on the x goes away, it's implied,
and then plus, 3 times 1 is 3,
and x^0, anything raised to the 0 power is just 1,
3 times 1 of course is just 3, that drops off.
And the answer here is 4x + 3.
Thanks. Bye!!!