DLS
  • DLS
Derivative help! If f(x)=cos(pi/2 [x]-x^3),1
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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DLS
  • DLS
\[\Huge f(x)=\cos(\frac{\pi}{2}[x]-x^3),1
DLS
  • DLS
\[\LARGE and~[x]=GINT \le x ,then~f'(\sqrt[3]{\frac{\pi}{2}})=?\]
DLS
  • DLS
@oldrin.bataku @Amber101 @mathslover @terenzreignz @Rayvenn.14

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DLS
  • DLS
@amistre64
anonymous
  • anonymous
Is [ ] referring to round, ceil, or floor function? @DLS
DLS
  • DLS
floor
anonymous
  • anonymous
Notice that the floor function [x] for 1 < x < 2, will always be equal to 1 hence mimicking the horizontal line x = 1. So the derivative with the Chain Rule becomes:\[\bf f'(x)=3x^2\sin \left( \frac{\pi}{2}-x^3 \right)\]
anonymous
  • anonymous
Notice that the [x] function will always equal to 1 for all 1 < x < 2. Hence we can eliminate it from the function itself and differentiate normally with the Chain Rule.
DLS
  • DLS
yo!
anonymous
  • anonymous
Now that we know the derivative, evaluate the derivative at the given x-value:\[\bf f'\left( \sqrt[3]{\frac{\pi}{2}} \right)=3\left( \sqrt[3]{\frac{\pi}{2}} \right)^2\sin\left( \frac{\pi}{2}-\frac{\pi}{2} \right)=0\] @DLS
anonymous
  • anonymous
@DLS Do you understand?
DLS
  • DLS
yes! thanks :D
anonymous
  • anonymous
$$f(x)=\cos\left(\frac\pi2\lfloor x\rfloor-x^3\right)$$Given \(1
anonymous
  • anonymous
@oldrin.bataku Problem has been solved already.
anonymous
  • anonymous
We could work the chain rule and simplify then or we could just let \(\lfloor x\rfloor =1\) now:$$f(x)=\cos\left(\frac\pi2-x^3\right)\\f'(x)=3x^2\sin\left(\frac\pi2-x^3\right)\\f'\left(\sqrt[3]{\frac\pi2}\right)=\left(\sqrt[3]{\frac\pi2}\right)^2\sin\left(\frac\pi2-\frac\pi2\right)=\sqrt[3]{\frac{\pi^2}{4}}\ \sin0=0$$
anonymous
  • anonymous
Alternatively,$$f(x)=\cos \left(\frac\pi2\lfloor x\rfloor-x^3\right)\\f'(x)=-\sin\left(\frac\pi2\lfloor x\rfloor-x^3\right)\left(\frac\pi2\frac{d}{dx}\lfloor x\rfloor-3x^2\right)$$We clearly find \(\frac{d}{dx}\lfloor x\rfloor =0\) everywhere it's defined (including \((1,2)\)) so simplify:$$f'(x)=3x^2\sin\left(\frac\pi2\lfloor x\rfloor-x^3\right)$$Clearly \(\lfloor x\rfloor=1\) for \(x\in(1,2)\) so:$$f'(x)=3x^2\sin\left(\frac\pi2-x^3\right)$$... etc.

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