How do I calculate the derivative using the chain and product rules? How can I calculate the derivative of the following function using the chain and product rules?
$y=30e^{-0.2x} \cdot \cos (1.5x) + 100$
I know I will have to use:
$y=vu'+uv'$
And I've found the answer using Wolfram Alpha - I just can't figure out the steps!
Thanks!
 A: 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 $$\frac{d}{dx}\cos (1.5x)=\frac{d}{d\left( 1.5x\right) }\cos (1.5x)\times 
\frac{d}{dx}\left( 1.5x\right) =(-\sin (1.5x))\left( 1.5\right). $$

