Question

Let F(x) = f(x ^ 9) and G(x) = (f(x)) ^ 9 You also know that a ^ 8 = 5 , f(a) = 2; f^ prime (a)=15, f^ prime (a^ 9 )=6

Answer 1

- The solution steps are given below;

Question 1

`\begin{gathered} F(x)=f(x^9) \\ \text{ Let }x^9=u \\ F(u)=f(u) \\ F^(\prime)(u)=(d)/(du)F(u)=f^(\prime)(u) \\ \\ (du)/(dx)=(d)/(dx)(x^9)=9x^8 \\ \\ F^(\prime)(x)=(d)/(du)F(u)*(du)/(dx) \\ \\ F^(\prime)(x)=f^(\prime)(u)*9x^8 \\ \\ F^(\prime)(x)=f^(\prime)(x^9)*9x^8 \\ \\ \text{ Thus, put }x=a \\ \\ F^(\prime)(a)=f^(\prime)(a^9)*9a^8 \\ \text{ But we have been given:} \\ f^(\prime)(a^9)=6 \\ a^8=5 \\ \\ \therefore F^(\prime)(a)=5*6=30 \end{gathered}`

Question 2:

`\begin{gathered} G(x)=(f(x))^9 \\ \text{ Let }u=f(x) \\ (d)/(dx)u=u^(\prime)=f^(\prime)(x) \\ \\ G(u)=u^9 \\ (d)/(du)G(u)=G^(\prime)(u)=9u^8 \\ \\ (d)/(du)G(u)*(d)/(dx)u=f^(\prime)(x)*9u^8 \\ \\ G^(\prime)(x)=f^(\prime)(x)*9(f(x))^8 \\ \\ put\text{ }x=a \\ \\ G^(\prime)(a)=f^(\prime)(a)*9(f(a))^8 \\ \text{ We know that:} \\ f^(\prime)(a)=15 \\ f(a)=2 \\ \\ \text{ Thus, we have:} \\ G^(\prime)(a)=15*9(2)^8 \\ G^(\prime)(a)=34560 \end{gathered}`