Question

Can you help me answer 20, 10 and 6 please

Answer 1

Let us the question 10. In this case

`z=(t^2+5t+2)/(t+3)`

Since the second degree polynomial cannott be factored, we must compute the derivative by means of the quotient rule

`z^(\prime)=((t+3)(2t+5)-(t^2+5t+2)(1))/((t+3)^2)`

Hence, we have

`z^(\prime)=(2t^2+5t+6t+15-(t^2+5t+2))/((t+3)^2)`

which is equal to

`z^(\prime)=(2t^2+5t+6t+15-t^2-5t-2))/((t+3)^2)`

It yields,

`z^(\prime)=(t^2+6t+13)/((t+3)^2)`

since the polynomial on top cannot be factored, therefore this is the final answer.

Let us see question 20. In this case

`f(x)=e^(-2x)\cdot\sin x`

By applying the chain rule, we have

`f^(\prime)(x)=e^(-2x)\cdot cosx+(-2)e^(-2x)\cdot\sin x`

Finally, this can be rewritten as

`f^(\prime)(x)=e^(-2x)(\cos x-2\sin x)`

Let us see question 6. In this case

`w=(t^3+5t)(t^2-7t+2)`

By applying the chain rule, we obtain

`w^(\prime)=(t^3+5t)(2t-7)+(3t^2+5)(t^2-7t+2)`

by computing the products, we have

`w^(\prime)=(2t^4-7t^3+10t-35)+(3t^4-21t^3+6t^2+5t^2-35t+10)`

hence, it yields

`w^(\prime)=5t^4-28t^3+11t^2-25t-25`