Derive the following function ( when solving you must use the rule of the product, the quotient rule and the chain rule): $$$f(x)=\frac{10e^{\sin(\sqrt[5]{x^2})} \cos(3x+1)}{\ln x}$$$
Development:
First, we must identify the different elementary functions, therefore we will have to use the quotient rule: $$$f'(x)=\dfrac{[10e^{\sin(\sqrt[5]{x^2})}\cos(3x+1)]' \cdot\ln(x)-[10e^{\sin(\sqrt[5]{x^2})}\cos(3x+1)]\dfrac{1}{x}}{\ln(x)^2}$$$
We will have to derive the following expression using the product rule:
$$[10e^{\sin(\sqrt[5]{x^2})}\cos(3x+1)]'=[10e^{\sin(\sqrt[5]{x^2})}]'\cos(3x+1)+10e^{\sin(\sqrt[5]{x^2})}[\cos(3x+1)]'$$
Let's keep on looking for the derivatives that we need by now using the chain rule:
$$[10e^{\sin(\sqrt[5]{x^2})}]'=10e^{\sin(\sqrt[5]{x^2})}\cos(\sqrt[5]{x^2})\dfrac{2}{5}x^{-3/5}=\dfrac{4e^{\sin(\sqrt[5]{x^2})}\cos(\sqrt[5]{x^2}) }{\sqrt[5]{x^3}} $$
$$[\cos(3x+1)]'=-\sin(3x+1)\cdot3=-3\sin(3x+1)$$
Introducing these results,
$$$f'(x)=\dfrac{[\dfrac{4e^{\sin(\sqrt[5]{x^2})}\cos(\sqrt[5]{x^2}) }{\sqrt[5]{x^3}}\cos(3x+1)-30\sin(3x+1)e^{\sin(\sqrt[5]{x^2})}] \cdot\ln(x)-[10e^{\sin(\sqrt[5]{x^2})}\cos(3x+1)]\dfrac{1}{x}}{\ln(x)^2}$$$
Solution:
$$$f'(x)=\dfrac{[\dfrac{4e^{\sin(\sqrt[5]{x^2})}\cos(\sqrt[5]{x^2}) }{\sqrt[5]{x^3}}\cos(3x+1)-30\sin(3x+1)e^{\sin(\sqrt[5]{x^2})}] \cdot\ln(x)-[10e^{\sin(\sqrt[5]{x^2})}\cos(3x+1)]\dfrac{1}{x}}{\ln(x)^2}$$$