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Derivative of $$$3 e^{- 4 r} \sin{\left(3 \theta \right)}$$$ with respect to $$$r$$$
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Your Input
Find $$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right)$$$.
Solution
Apply the constant multiple rule $$$\frac{d}{dr} \left(c f{\left(r \right)}\right) = c \frac{d}{dr} \left(f{\left(r \right)}\right)$$$ with $$$c = 3 \sin{\left(3 \theta \right)}$$$ and $$$f{\left(r \right)} = e^{- 4 r}$$$:
$${\color{red}\left(\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right)\right)} = {\color{red}\left(3 \sin{\left(3 \theta \right)} \frac{d}{dr} \left(e^{- 4 r}\right)\right)}$$The function $$$e^{- 4 r}$$$ is the composition $$$f{\left(g{\left(r \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = e^{u}$$$ and $$$g{\left(r \right)} = - 4 r$$$.
Apply the chain rule $$$\frac{d}{dr} \left(f{\left(g{\left(r \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dr} \left(g{\left(r \right)}\right)$$$:
$$3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(e^{- 4 r}\right)\right)} = 3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dr} \left(- 4 r\right)\right)}$$The derivative of the exponential is $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$$3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dr} \left(- 4 r\right) = 3 \sin{\left(3 \theta \right)} {\color{red}\left(e^{u}\right)} \frac{d}{dr} \left(- 4 r\right)$$Return to the old variable:
$$3 e^{{\color{red}\left(u\right)}} \sin{\left(3 \theta \right)} \frac{d}{dr} \left(- 4 r\right) = 3 e^{{\color{red}\left(- 4 r\right)}} \sin{\left(3 \theta \right)} \frac{d}{dr} \left(- 4 r\right)$$Apply the constant multiple rule $$$\frac{d}{dr} \left(c f{\left(r \right)}\right) = c \frac{d}{dr} \left(f{\left(r \right)}\right)$$$ with $$$c = -4$$$ and $$$f{\left(r \right)} = r$$$:
$$3 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(- 4 r\right)\right)} = 3 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(- 4 \frac{d}{dr} \left(r\right)\right)}$$Apply the power rule $$$\frac{d}{dr} \left(r^{n}\right) = n r^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dr} \left(r\right) = 1$$$:
$$- 12 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(r\right)\right)} = - 12 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(1\right)}$$Thus, $$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right) = - 12 e^{- 4 r} \sin{\left(3 \theta \right)}$$$.
Answer
$$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right) = - 12 e^{- 4 r} \sin{\left(3 \theta \right)}$$$A