calculate the $(f(g(x)))$ and $(g(f(x)))$ derivatives if\n\n$f(x)=x^{6}$\n\n$g(x)=\frac{1}{7x +…

Answer:

1. $(f(g(x)))'=-\frac{6}{(7x + 1)^{7}}$
2. $(g(f(x)))'=-\frac{42x^{5}}{(7x^{6}+1)^{2}}$

Explanation:

Step1: Find $f(g(x))$

$f(g(x))=\left(\frac{1}{7x + 1}\right)^{6}=(7x + 1)^{-6}$

Step2: Differentiate $f(g(x))$ using chain - rule

Let $u = 7x+1$, then $y = u^{-6}$. $\frac{dy}{du}=-6u^{-7}$ and $\frac{du}{dx}=7$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. So $(f(g(x)))'=-6(7x + 1)^{-7}\cdot7=-\frac{6}{(7x + 1)^{7}}$

Step3: Find $g(f(x))$

$g(f(x))=\frac{1}{7x^{6}+1}=(7x^{6}+1)^{-1}$

Step4: Differentiate $g(f(x))$ using chain - rule

Let $u = 7x^{6}+1$, then $y = u^{-1}$. $\frac{dy}{du}=-u^{-2}$ and $\frac{du}{dx}=42x^{5}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. So $(g(f(x)))'=- (7x^{6}+1)^{-2}\cdot42x^{5}=-\frac{42x^{5}}{(7x^{6}+1)^{2}}$