Thread: First-Order Derivative

I'm not sure if i'm doing this right, it's been awhile since i've had to deal with calculus!
So it says calculate the first-order derivative with respect to x:

a. y = 3x^2
Ans: F(x) = 9X^2

b. y = x^3
Ans: F(x) 3x^2

c. y=mx + b
Ans: Not sure on this one ..

d. y = 1/x
Ans: Also not sure

e. y = 37x^4
Ans: F(x) = 148x^3

f. y = x^a/b
ANS: F(x) = a/b x^a/b - 1

Okay. maybe i'm completely wrong with some of these!
any help is greatly appreciated!

a. y = 3x^2
Ans: F(x) = 9X^2

Recheck.

Originally Posted by jcrane

c. y=mx + b
Ans: Not sure on this one ..

$\displaystyle m$ and $\displaystyle b$ are constant, just like any other.

d. y = 1/x
Ans: Also not sure

$\displaystyle \frac{1}{x} = x^{-1}$.

Recall $\displaystyle y= x^n \implies \frac{dy}{dx}= nx^{n-1}$

Originally Posted by jcrane

a. y = 3x^2
Ans: F(x) = 9X^2

Therefore

$\displaystyle \frac{dy}{dx}=9x$

Originally Posted by jcrane

b. y = x^3
Ans: F(x) 3x^2

Yep

Originally Posted by jcrane

c. y=mx + b
Ans: Not sure on this one ..

treat m and b as a constant,

Originally Posted by jcrane

d. y = 1/x
Ans: Also not sure

$\displaystyle \frac{1}{x}= x^{-1}$

Err...pickslides, I think you made a "silly mistake" on the first one.

$\displaystyle \frac{d}{dx}\left(3x^2\right) = 2\cdot 3x^{2-1}\neq 9x$