Find derivatives

If f(x)=exsinxf(x) = e^x \sin{x}, then d10dx10f(x)\frac{\text{d}^{10}}{\text{d} x^{10}} f(x) at x=0x=0 equals:

  • 1

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  • 10

  • 32.

Can you generalise this?

#Calculus #Derivatives #Goldbach'sConjurersGroup

Note by A Brilliant Member
7 years, 4 months ago

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Comments

Sure. f(x)=exsinx=(ex(cosx+isinx))=(exeix)=(ex(1+i))f(x)=e^x\sin x=\Im(e^x(\cos x+i\sin x))=\Im(e^xe^{ix})=\Im(e^{x(1+i)}). Hence, dnfdxn=(dndxne(1+i)x)=((1+i)ne(1+i)x)\frac{d^{n}f}{dx^n}=\Im(\frac{d^n}{dx^n}e^{(1+i)x})=\Im((1+i)^ne^{(1+i)x}). We can rewrite (1+i)n=2n/2einπ4=2n/2(cosnπ4+isinnπ4)(1+i)^n=2^{n/2}e^{i\frac{n\pi}{4}}=2^{n/2}(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}). When multiplied, we get e(1+i)x=1e^{(1+i)x}=1 at x=0x=0, so dnfdxn=(2n/2(cosnπ4+isinnπ4))=2n/2sinnπ4\frac{d^nf}{dx^n}=\Im(2^{n/2}(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}))=\boxed{2^{n/2}\sin\frac{n\pi}{4}}.

Cody Johnson - 7 years, 4 months ago

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Lovely!

A Brilliant Member - 7 years, 4 months ago

32

Carlos Suarez - 7 years, 1 month ago
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