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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:


And I've found the answer using Wolfram Alpha - I just can't figure out the steps!


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There is a misprint in your post: the fact you know you will have to use is $(uv)'=u'v+uv'$. – Did Aug 11 '11 at 12:04
There is a link on right upper part in WolframAlpha which says show steps. Can come in handy once in a while ;) EDIT: But sometimes it doesn't show up. Type the following in Wolfram Alpha d[30*Exp[-0.2x]*Cos[3x/2] + 100]/dx – user13838 Aug 11 '11 at 12:15
up vote 3 down vote accepted

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).

$$\begin{eqnarray*} y^{\prime } &=&\frac{d}{dx}\left( 30e^{-0.2x}\cos (1.5x)+83.4\right) \\ &=&\frac{d}{dx}\left( 30e^{-0.2x}\cos (1.5x)\right) +\frac{d}{dx}83.4\qquad\qquad\qquad\qquad\qquad\qquad\text{by rule 1} \\ &=&30\frac{d}{dx}\left( e^{-0.2x}\cos (1.5x)\right) +0 \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{ by rules 4 and 3} \\ &=&30\left( \left( \frac{d}{dx}e^{-0.2x}\right) \cos (1.5x)+e^{-0.2x}\frac{d% }{dx}\cos (1.5x)\right)\qquad\text{by rule 2} \\ &=&30\left( \left( e^{-0.2x}\frac{d}{dx}\left( -0.2x\right) \right) \cos (1.5x)+e^{-0.2x}\left( -\sin (1.5x)\right) \frac{d}{dx}(1.5x)\right)\quad\text{by rule 5} \\ &=&30\left( \left( e^{-0.2x}\left( -0.2\right) \right) \cos (1.5x)+e^{-0.2x}\left( -\sin (1.5x)\right) (1.5)\right) \\ &=&-30\left( 0.2e^{-0.2x}\cos (1.5x)+1.5e^{-0.2x}\sin (1.5x)\right) \\ &=&-6e^{-0.2x}\cos 1.5x-45e^{-0.2x}\sin \left( 1.5x\right) \end{eqnarray*}$$


  1. Application of the sum rule: $$(u+v)'=u'+v'\qquad(1)$$ to $\frac{d}{dx}\left( 30e^{-0.2x}\cos (1.5x)+83.4\right)$: $$\frac{d}{dx}\left( 30e^{-0.2x}\cos (1.5x)+83.4\right)=\frac{d}{dx}\left( 30e^{-0.2x}\cos (1.5x)\right) +\frac{d}{dx}83.4$$
  2. 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 $$\frac{d}{dx}\left( e^{-0.2x}\cos (1.5x)\right) =\left( \frac{d}{dx}% e^{-0.2x}\right) \cos (1.5x)+e^{-0.2x}\frac{d}{dx}\cos (1.5x).$$
  3. Derivative of a constant $c$: $$\frac{d}{dx}c=0.\qquad(3)$$
  4. Derivative of a product of a constant with a function (particular case of $(2)$): $$\frac{d}{dx}cf(x)=c\frac{d}{dx}f(x)=cf'(x).\qquad(4)$$
  5. 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 $$\frac{d}{dx}\cos (1.5x)=\frac{d}{d\left( 1.5x\right) }\cos (1.5x)\times \frac{d}{dx}\left( 1.5x\right) =(-\sin (1.5x))\left( 1.5\right). $$
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