Basic Derivatives Example 2


Uploaded by TheIntegralCALC on 12.02.2010

Transcript:
Hey everyone!
What's up?
Another derivative problem,
this one is slightly more complicated,
but only slightly,
f(x) = 100 - 16x^2.
The last derivative problem we did didn't have an exponent, now we've got an exponent.
So, derivative notation, f'(x)
equals...
As we've learned last time the derivative of a constant is 0,
so that goes away,
and then, all we have to deal with is this 16x^2 here.
We'll also have to keep in mind that there's a... there's a negative in front of it.
So the way that we're going to take the derivative of this,
as before, is the exponent times the coefficient.
So it's going to be 2 times 16.
Oh, I can see I already made a mistake, I... I forgot the negative sign,
so, we have to put the negative sign in front of it here,
and then the exponent times the coefficient, 2 times 16,
the x we keep, and we subtract 1 from the exponent,
so 2 - 1 is 1.
And then, we just simplify this,
the derivative is negative,
2 times 16, 32,
x, and then we can just drop the 1 obviously as the exponent,
so -32x here is our answer.
See you later!