Let us the question 10. In this case
[tex]z=\frac{t^2+5t+2}{t+3}[/tex]Since the second degree polynomial cannott be factored, we must compute the derivative by means of the quotient rule
[tex]z^{\prime}=\frac{(t+3)(2t+5)-(t^2+5t+2)(1)}{(t+3)^2}[/tex]Hence, we have
[tex]z^{\prime}=\frac{2t^2+5t+6t+15-(t^2+5t+2)}{(t+3)^2}[/tex]which is equal to
[tex]z^{\prime}=\frac{2t^2+5t+6t+15-t^2-5t-2)}{(t+3)^2}[/tex]It yields,
[tex]z^{\prime}=\frac{t^2+6t+13}{(t+3)^2}[/tex]since the polynomial on top cannot be factored, therefore this is the final answer.
Let us see question 20. In this case
[tex]f(x)=e^{-2x}\cdot\sin x[/tex]By applying the chain rule, we have
[tex]f^{\prime}(x)=e^{-2x}\cdot cosx+(-2)e^{-2x}\cdot\sin x[/tex]Finally, this can be rewritten as
[tex]f^{\prime}(x)=e^{-2x}(\cos x-2\sin x)[/tex]Let us see question 6. In this case
[tex]w=(t^3+5t)(t^2-7t+2)[/tex]By applying the chain rule, we obtain
[tex]w^{\prime}=(t^3+5t)(2t-7)+(3t^2+5)(t^2-7t+2)[/tex]by computing the products, we have
[tex]w^{\prime}=(2t^4-7t^3+10t-35)+(3t^4-21t^3+6t^2+5t^2-35t+10)[/tex]hence, it yields
[tex]w^{\prime}=5t^4-28t^3+11t^2-25t-25[/tex]