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

Solution

- The solution steps are given below;

Question 1

$\begin{gathered} F(x)=f(x^9) \\ \text{ Let }x^9=u \\ F(u)=f(u) \\ F^{\prime}(u)=\frac{d}{du}F(u)=f^{\prime}(u) \\ \\ \frac{du}{dx}=\frac{d}{dx}(x^9)=9x^8 \\ \\ F^{\prime}(x)=\frac{d}{du}F(u)\times\frac{du}{dx} \\ \\ F^{\prime}(x)=f^{\prime}(u)\times9x^8 \\ \\ F^{\prime}(x)=f^{\prime}(x^9)\times9x^8 \\ \\ \text{ Thus, put }x=a \\ \\ F^{\prime}(a)=f^{\prime}(a^9)\times9a^8 \\ \text{ But we have been given:} \\ f^{\prime}(a^9)=6 \\ a^8=5 \\ \\ \therefore F^{\prime}(a)=5\times6=30 \end{gathered}$

Question 2:

$\begin{gathered} G(x)=(f(x))^9 \\ \text{ Let }u=f(x) \\ \frac{d}{dx}u=u^{\prime}=f^{\prime}(x) \\ \\ G(u)=u^9 \\ \frac{d}{du}G(u)=G^{\prime}(u)=9u^8 \\ \\ \frac{d}{du}G(u)\times\frac{d}{dx}u=f^{\prime}(x)\times9u^8 \\ \\ G^{\prime}(x)=f^{\prime}(x)\times9(f(x))^8 \\ \\ put\text{ }x=a \\ \\ G^{\prime}(a)=f^{\prime}(a)\times9(f(a))^8 \\ \text{ We know that:} \\ f^{\prime}(a)=15 \\ f(a)=2 \\ \\ \text{ Thus, we have:} \\ G^{\prime}(a)=15\times9(2)^8 \\ G^{\prime}(a)=34560 \end{gathered}$