1) $\displaystyle f(x) = (x-3)x^\frac{1}{2}$ <--- to the one half power.
2) $\displaystyle g(x) = (x^2-3x)^-2 + 2$ <---- to the -2 power plus 2
3) $\displaystyle h(x) = (9-x^2)^2$
Thanks for any help.
1) $\displaystyle f(x) = (x-3)x^\frac{1}{2}$ <--- to the one half power.
2) $\displaystyle g(x) = (x^2-3x)^-2 + 2$ <---- to the -2 power plus 2
3) $\displaystyle h(x) = (9-x^2)^2$
Thanks for any help.
1) $\displaystyle f(x) = (x^1-3)x^{\frac{1}{2}}$
$\displaystyle f(x) =x^{\frac{3}{2}}-3x^{\frac{1}{2}}$
Now, differentiate, got it ???
2) $\displaystyle g(x) = \frac{1}{(x^2+3x)^2}+2$
$\displaystyle g(x) = \frac{1}{x^4+6x^3+9x^2}+2$
Now, differentiate, using Quotient Rule, got it ???
3) $\displaystyle h(x) = (9-x^2)^2$
$\displaystyle
h(x) = 81-18x^2+x^4$
Now, differentiate, got it ???
I think I finished it... but I'm not sure if I did it right.. Can someone check for me?
1. $\displaystyle {\frac{3}{2}}x^{\frac{1}{2}}-{\frac{3}{2}}x^{\frac{-1}{2}}$
For number 2, I have a problem. After I used the quotient rule, I came upon
$\displaystyle -4x^3-18x^2-18x/(x^4+6x^3+9x^2)^2$ and I don't know how to simplify that.
yes
why use the quotient rule? as the problem was written, it seems to suggest you should use the chain rule. much in the same way you did problem 3For number 2, I have a problem. After I used the quotient rule, I came upon
$\displaystyle -4x^3-18x^2-18x/(x^4+6x^3+9x^2)^2$ and I don't know how to simplify that.
Yes, the chain rule is easier for 2).
$\displaystyle g = (x^2 - 3x)^{-2} + 2. $
Let $\displaystyle u = x^2 - 3x$ so $\displaystyle g = u^{-2} + 2.$
$\displaystyle \frac{du}{dx} = 2x, \frac{dg}{du} = -2u^{-3} = -2(x^2 - 3x)^{-3}$
So $\displaystyle \frac{dg}{dx} = \frac{du}{dx}\times\frac{dg}{du} = -4x(x^2 - 3x)^{-3}$.