1. ## derivative check

f(x) = (sqrt(x))/(x^3+1)
f'(x) = (-2.5x^5/2)+(1/2sqrt(x)) / (x^3+1)^2

f(x) = (cosx/t^3)
f'(x) = (-x^3sinx)-(3x²cosx)/(t^3)²

f(x) = [2-(1/x)]/(x-3)
f'(x) = (-2x²+2x-3)/(x²-3x)²

unsuree if i have done these correctly on my H/W...-thx

2. Originally Posted by maybnxtseasn
f(x) = (sqrt(x))/(x^3+1)
f'(x) = (-2.5x^5/2)+(1/2sqrt(x)) / (x^3+1)^2

f(x) = (cosx/t^3)
f'(x) = (-x^3sinx)-(3x²cosx)/(t^3)²

f(x) = [2-(1/x)]/(x-3)
f'(x) = (-2x²+2x-3)/(x²-3x)²

unsuree if i have done these correctly on my H/W...-thx
Here's a kickstart on the first one:

$\displaystyle f(x) = \frac{\sqrt x}{x^3+1}$

$\displaystyle f'(x) = \frac {(x^3+1) \frac{1}{2}x^{-\frac{1}{2}} - \sqrt{x} \cdot 3x^2}{(x^3+1)^2} = \frac {\frac{1}{2}x^{\frac{5}{2}} + \frac{1}{2}x^{-\frac{1}{2}} - 3x^{\frac{5}{2}}}{(x^3+1)^2} = \frac {-\frac{5}{2}x^{\frac{5}{2}} + \frac{1}{2}x^{-\frac{1}{2}}}{(x^3+1)^2}$

I can't read your Math clearly, but I think you got it right. By
(1/2sqrt(x))
, you mean $\displaystyle \frac{1}{2\sqrt{x}}$ right? Then, I think what you have is correct.

3. Originally Posted by maybnxtseasn
f(x) = (sqrt(x))/(x^3+1)
f'(x) = (-2.5x^5/2)+(1/2sqrt(x)) / (x^3+1)^2

f(x) = (cosx/t^3)
f'(x) = (-x^3sinx)-(3x²cosx)/(t^3)²

f(x) = [2-(1/x)]/(x-3)
f'(x) = (-2x²+2x-3)/(x²-3x)²

unsuree if i have done these correctly on my H/W...-thx
And for the second one:
$\displaystyle f(x) = \frac{cosx}{t^3} = t^{-3}cosx$

You are forgetting implicit differentiation here. You are differentiating with respect to x, so when you take the derivative of t it will be t' or $\displaystyle \frac {dt}{dx}$, but not 1.

$\displaystyle f'(x) = -t^{-3}sinx - 3t^{-4}cosx \cdot t'$

You also seem to be changing the t into an x sometimes. Make sure you understand what I did there. If you want me to explain it in more detail, I will.
Hope I helped you

4. Hello, maybnxtseasn!

$\displaystyle f(x) \:= \:\frac{x^{\frac{1}{2}}} {x^3+1}$

$\displaystyle f'(x) \:=\: \frac{-2.5x^{\frac{5}{2}} +{\color{red}\dfrac{1}{2\sqrt{x}}}}{(x^3+1)^2}$ . . Is this what you meant?

$\displaystyle f'(x) \;=\;\frac{(x^3+1)\cdot \frac{1}{2}x^{-\frac{1}{2}} - x^{\frac{1}{2}}(3x^2)}{(x^3+1)^2} \;=\;\frac{\frac{1}{2}x^{-\frac{1}{2}}(x^3+1) - 3x^{\frac{5}{2}}}{(x^3+1)^2}$

Multiply by $\displaystyle \frac{2x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}$

.$\displaystyle f'(x) \;=\;\frac{2x^{\frac{1}{2}} \bigg[ \frac{1}{2}x^{-\frac{1}{2}}(x^3+1) - 3x^{\frac{5}{2}} \bigg]} {2x^{\frac{1}{2}}\cdot (x^3+1)^2}$

. . . . $\displaystyle =\;\frac{x^3+1 - 6x^3}{2x^{\frac{1}{2}}(x^3+1)^2} \;=\;\frac{-5x^3 + 1}{2\sqrt{x}\,(x^3+1)^2}$

. . which is equivalent to your result.

5. ## yo

Originally Posted by mathemagister
And for the second one:
$\displaystyle f(x) = \frac{cosx}{t^3} = t^{-3}cosx$

You are forgetting implicit differentiation here. You are differentiating with respect to x, so when you take the derivative of t it will be t' or $\displaystyle \frac {dt}{dx}$, but not 1.

$\displaystyle f'(x) = -t^{-3}sinx - 3t^{-4}cosx \cdot t'$

You also seem to be changing the t into an x sometimes. Make sure you understand what I did there. If you want me to explain it in more detail, I will.
Hope I helped you
i still dont see how u got that awnser....anyone else help? how come i dont get this awnser when i do the simple quotient rule with this?