Problems from Derivative of the composition of functions (chain rule)

Development:

First, we must identify the different elementary functions, therefore we will have to use the quotient rule: $$f^{\prime }(x)={\displaystyle \frac{[10e^{\sin (\sqrt[5]{x^2})}\cos (3x+1){]}^{\prime }\cdot \ln (x)-[10e^{\sin (\sqrt[5]{x^2})}\cos (3x+1)]{\displaystyle \frac{1}{x}}}{\ln (x{)}^2}}$$

We will have to derive the following expression using the product rule:

$[10e^{\sin (\sqrt[5]{x^2})}\cos (3x+1){]}^{\prime }=[10e^{\sin (\sqrt[5]{x^2})}{]}^{\prime }\cos (3x+1)+10e^{\sin (\sqrt[5]{x^2})}[\cos (3x+1){]}^{\prime }$

Let's keep on looking for the derivatives that we need by now using the chain rule:

$[10e^{\sin (\sqrt[5]{x^2})}{]}^{\prime }=10e^{\sin (\sqrt[5]{x^2})}\cos (\sqrt[5]{x^2}){\displaystyle \frac{2}{5}}{x}^{-3/5}={\displaystyle \frac{4e^{\sin (\sqrt[5]{x^2})}\cos (\sqrt[5]{x^2})}{\sqrt[5]{x^3}}}$

$[\cos (3x+1){]}^{\prime }=-\sin (3x+1)\cdot 3=-3\sin (3x+1)$

Introducing these results,

$$f^{\prime }(x)={\displaystyle \frac{[{\displaystyle \frac{4e^{\sin (\sqrt[5]{x^2})}\cos (\sqrt[5]{x^2})}{\sqrt[5]{x^3}}}\cos (3x+1)-30\sin (3x+1)e^{\sin (\sqrt[5]{x^2})}]\cdot \ln (x)-[10e^{\sin (\sqrt[5]{x^2})}\cos (3x+1)]{\displaystyle \frac{1}{x}}}{\ln (x{)}^2}}$$

Solution:

$$f^{\prime }(x)={\displaystyle \frac{[{\displaystyle \frac{4e^{\sin (\sqrt[5]{x^2})}\cos (\sqrt[5]{x^2})}{\sqrt[5]{x^3}}}\cos (3x+1)-30\sin (3x+1)e^{\sin (\sqrt[5]{x^2})}]\cdot \ln (x)-[10e^{\sin (\sqrt[5]{x^2})}\cos (3x+1)]{\displaystyle \frac{1}{x}}}{\ln (x{)}^2}}$$

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