Differentiate

1. f(x)= x^5(cosx)

2. u= (2x+1)5^x

3. g(s)= s^(5/3) - 2se^s
My answer: 5s^(2/3)/3 - 2e^s + 2se^s

4. y= 1/(x^3 + 3x + 1)
My answer: 3x^2+3/(x^3 + 3x +1)^2

5. y= 4/(e^t + 1)

6. w= x/(x^2 + 2)
My answer: -x^3 + 2x^2 -2x/ (x^2 + 2)^2

2. Originally Posted by jkami
Differentiate

1. f(x)= x^5(cosx)
Correct.

2. u= (2x+1)5^x
Correct.

3. g(s)= s^(5/3) - 2se^s
My answer: 5s^(2/3)/3 - 2e^s + 2se^s
Correct, only the final term must be $\displaystyle -2se^s$.

4. y= 1/(x^3 + 3x + 1)
My answer: 3x^2+3/(x^3 + 3x +1)^2
Remember that $\displaystyle \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}$.

5. y= 4/(e^t + 1)
Correct.

6. w= x/(x^2 + 2)
My answer: -x^3 + 2x^2 -2x/ (x^2 + 2)^2
I derived:

\displaystyle \begin{aligned} \frac{dw}{dx}&=\frac{d}{dx}\left(\frac{x}{x^2+2}\r ight)\\ &=\frac{(x^2+2)\cdot 1 - x\cdot 2x}{(x^2+2)^2}\\ &=\frac{x^2+2-2x^2}{(x^2+2)^2}\\ &=\frac{-x^2+2}{(x^2+2)^2}. \end{aligned}

3. You have an error when you differentiate the number 4, it must be:
$\displaystyle y'=\dfrac{-(3x^2+3)}{(x^3 + 3x +1)^2}$.
Also in the number 6:
$\displaystyle y'=\dfrac{\{x\}'(x^2+2)-x\{x^2+2\}'}{(x^2 + 2)^2}=\dfrac{x^2+2-2x^2}{(x^2 + 2)^2}=\dfrac{2-x^2}{(x^2+2)^2}$

4. Another way to differentiate last question could be

$\displaystyle w(x)=\frac{x}{{{x}^{2}}+2}=\frac{1}{x+2{{x}^{-1}}}={{\left( x+2{{x}^{-1}} \right)}^{-1}},$ then $\displaystyle w'(x)=-\left( 1-\frac{2}{{{x}^{2}}} \right)\cdot \frac{1}{{{\left( x+2{{x}^{-1}} \right)}^{2}}}=\frac{2-{{x}^{2}}}{{{\left( {{x}^{2}}+2 \right)}^{2}}}.$