Here is the detailed evaluation. It uses the rules of the derivatives. The sum rule, the rule of a product of a constant with a function, the product rule, the chain rule and the derivative of a constant. (See details below).
$\begin{eqnarray*} y^{\prime } &=&\frac{d}{dx}\left( 30e^{-0.2x}\cos (1.5x)+83.4\right) \\ &=&\frac{d}{dx}\left( 30e^{-0.2x}\cos (1.5x)\right) +\frac{d}{dx}83.4\qquad\qquad\qquad\qquad\qquad\qquad\text{by rule 1} \\ &=&30\frac{d}{dx}\left( e^{-0.2x}\cos (1.5x)\right) +0 \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{ by rules 4 and 3} \\ &=&30\left( \left( \frac{d}{dx}e^{-0.2x}\right) \cos (1.5x)+e^{-0.2x}\frac{d% }{dx}\cos (1.5x)\right)\qquad\text{by rule 2} \\ &=&30\left( \left( e^{-0.2x}\frac{d}{dx}\left( -0.2x\right) \right) \cos (1.5x)+e^{-0.2x}\left( -\sin (1.5x)\right) \frac{d}{dx}(1.5x)\right)\quad\text{by rule 5} \\ &=&30\left( \left( e^{-0.2x}\left( -0.2\right) \right) \cos (1.5x)+e^{-0.2x}\left( -\sin (1.5x)\right) (1.5)\right) \\ &=&-30\left( 0.2e^{-0.2x}\cos (1.5x)+1.5e^{-0.2x}\sin (1.5x)\right) \\ &=&-6e^{-0.2x}\cos 1.5x-45e^{-0.2x}\sin \left( 1.5x\right) \end{eqnarray*}$
Rules:
- Application of the sum rule: (u+v)'=u'+v'\qquad(1) to $\frac{d}{dx}\left( 30e^{-0.2x}\cos (1.5x)+83.4\right)$: $\frac{d}{dx}\left( 30e^{-0.2x}\cos (1.5x)+83.4\right)=\frac{d}{dx}\left( 30e^{-0.2x}\cos (1.5x)\right) +\frac{d}{dx}83.4$
- Application of the product rule: $y^{\prime }=\left( uv\right) ^{\prime }=u^{\prime }v+uv^{\prime },\qquad (2)$ with $u=e^{-0.2x}$, $v=\cos (1.5x)$. Thus $u^{\prime }=\left( e^{-0.2x}\right) ^{\prime }$, $v^{\prime }=\left( \cos (1.5x)\right) ^{\prime }$ and $\left( uv\right) ^{\prime }=\left( e^{-0.2x}\right) ^{\prime }\cos (1.5x)+e^{-0.2x}\left( \cos (1.5x)\right) ^{\prime }$ or $\frac{d}{dx}\left( e^{-0.2x}\cos (1.5x)\right) =\left( \frac{d}{dx}% e^{-0.2x}\right) \cos (1.5x)+e^{-0.2x}\frac{d}{dx}\cos (1.5x).$
- Derivative of a constant $c$: $\frac{d}{dx}c=0.\qquad(3)$
- Derivative of a product of a constant with a function (particular case of $(2)$): \frac{d}{dx}cf(x)=c\frac{d}{dx}f(x)=cf'(x).\qquad(4)
- Application of the chain rule: if $u=g(x)$, then $\frac{d}{dx}f(g(x))=\left( \frac{df(u)}{du}\right) _{u=g(x)}\times \frac{% dg(x)}{dx}.\qquad(5)$ For $u=g(x)=-0.2x$, $f(g(x))=e^{-0.2x}$, f(u)=e^{u}=f'(u), $u^{\prime }=g^{\prime }(x)=-0.2$. Hence, in a different notation: $\frac{d}{dx}e^{-0.2x}=\frac{d}{d\left( -0.2x\right) }\left( e^{-0.2x}\right) \times \frac{d}{dx}\left( -0.2x\right) =e^{-0.2x}\left( -0.2\right) $ Similarly, we get $\frac{d}{dx}\cos (1.5x)=\frac{d}{d\left( 1.5x\right) }\cos (1.5x)\times \frac{d}{dx}\left( 1.5x\right) =(-\sin (1.5x))\left( 1.5\right). $