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, you mean $\displaystyle \frac{1}{2\sqrt{x}}$ right? Then, I think what you have is correct.(1/2sqrt(x))
Hope that was helpful
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
Hello, maybnxtseasn!
I think I agree with your first answer . . .
$\displaystyle f(x) \:= \:\frac{x^{\frac{1}{2}}} {x^3+1}$
$\displaystyle f'(x) \:=\: \frac{-2.5x^{\frac{5}{2}} +{\color{red}\dfrac{1}{2\sqrt{x}}}}{(x^3+1)^2}$ . . 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.