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I am trying to calculate the derivative of a rather complex function for my homework. I think I have found the solution, it just seems too bulky for my taste. See the bottom for specific questions I have regarding my solution.

$ f(x)=\frac{\overbrace{\sin x}^\text{u(x)}\cdot \overbrace{e^x+x^3}^\text{v(x)}}{\underbrace{x^3+2x+2}_\text{w(x)}} $

$ u'(x)=\cos x $, $ v'(x)=e^x+3x^2 $, $ w'(x)=3x^2+4x $

$ \begin{align*} u'v'(x)&=(\sin x\cdot e^x+3x^2)+(\cos x\cdot x^3)\\ &= e^x\cdot\sin x+3x\cdot\sin x+x^3\cdot\cos x \end{align*} $

$ \begin{align*} f'(x)&= \frac{u'v'(x)\cdot w(x)-w'(x)\cdot uv(x)}{w(x)^2}\\ &= \frac{(e^x\cdot\sin x+3x\cdot\sin x+x^3\cdot\cos x)\cdot (x^3+2x^2+2)-(3x^2+4x)(\sin x\cdot e^x+x^3)}{(x^3+2x^2+2)^2} \end{align*} $

The obvious question: Is this derivative correct?

Especially:

  • Is it really possible to calculate the derivative by splitting the function into part-functions and calculating them together following the differentiation rules, the way I did it?

  • Is it possible to reduce the summands by applying some rule I am not aware of? E.g. reducing $(e^x\cdot\sin x+3x\cdot\sin x+x^3\cdot\cos x)$ to something with just one $\sin$ or something?

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  • $\begingroup$ $\sin x\cdot e^x+x^3=e^x\sin x+x^3$, not $(e^x+x^3)\sin x$. $\endgroup$ Jan 3, 2013 at 19:53
  • $\begingroup$ Students use period '.' instead of parentheses. It is not standard though. So it is not clear what is in the numerator. On the second line you mean to write $(uv)'=(\sin x . e^x + 3x^2)+ ...$. Same error show on the line $f'(x)=...$. Otherwise it looks OK to me. You can check your answers on many sites these days. Here is one $\endgroup$
    – Maesumi
    Jan 3, 2013 at 20:09
  • $\begingroup$ This was what I was thinking. However, because of the product rule I thought I had to actually calculate the part-functions first, i.e. the derivatives of $a(x)=\sin x$, $b(x)=eˆx$ and $c(x)=xˆ3$ (with $v(x)=a(x)+b(x)$). If that is not necessary, then @dirk5959's answer (using simply the quotient-rule) is of course correct and evident! $\endgroup$
    – alex
    Jan 3, 2013 at 21:31
  • $\begingroup$ If the expression is meant as $uv/w$ then use $(uv)'=uv'+u' v=\sin x (e^x+3x^2) + \cos x (e^x+x^3) $. As as there are two errors. You have only $\cos x . x^3$ missing an $e^x$ there. Next you have written $3x$ instead of $3x^2$. $\endgroup$
    – Maesumi
    Jan 3, 2013 at 21:43

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Your solution is pretty close, but remember that you're not multiplying $\sin x$ by $(e^x + x^3)$ in your numerator, as you do when you try to multiply $u'(x)$ by $v'(x).$ Instead, try differentiating your numerator in the following form: $e^x\sin x + x^3.$ I believe the rest will follow from the quotient rule.

$n(x) = e^x\sin x$

$d(x) = x^3 + 2x + 2$

$n'(x) = e^x \cos x + e^x\sin x$

$d'(x) = 3x^2 + 2$

$\frac{d}{dx} \frac{n(x)}{d(x)} = \frac{n'(x)d(x) - d'(x)n(x)}{d(x)^2}$

$ = \frac{(e^x\cos x + e^x\sin x)(x^3 + 2x + 2) - (3x^2+2)(e^x\sin x)}{(x^3 + 2x + 2)^2}$

which is messy but can be simplified.

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