+14905 Was ist die 101. Ableitung von f(x)=cos(x)?
\(\ f(x)\ =cos(x),\ f'(x)\ =-sin(x),\ f''(x)\ = -cos(x), f^{3'}(x)=sin(x)\)
\(f^{4'}(x)=cos(x),\ f^{5'}(x)=-sin(x),\ f^{6'}(x)=-cos(x), f^{7'}(x)=sin(x)\)
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\(\ f^{96'}(x)=\) .. .. \(f^{99'}(x)=sin(x)\)
\(f^{100'}(x)=cos(x),\ f^{101'}(x)=-sin(x),\ f^{102'}(x)=-cos(x), f^{103'}(x)=sin(x)\)
\(n\in \mathbb{N}\\ f(x)=cos(x)\\ f^{n'}=cos(x)\ |\frac{n}{4}\in \mathbb{N}\\ f^{n'}=-sin(x)\ |\frac{n+3}{4}\in \mathbb{N}\\ f^{n'}=-cos(x)\ |\frac{n+2}{4}\in \mathbb{N}\\ f^{n'}=sin(x)\ |\frac{n+1}{4}\in \mathbb{N}\\\)
\(f^{101'}=-sin(x)\ |(\frac{101+3}{4}=26)\in \mathbb{N}\)
Die 101. Ableitung von f(x) = cos(x)
ist
\(f^{101'}(x)=-sin(x)\)
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+26367 Was ist die 101. Ableitung von \(f(x)=\cos(x)\)?
\(\begin{array}{|rcll|} \hline f(x) &=& \cos(x) \\ &=& \cos\left(0\cdot \dfrac{\pi}{2} + x \right) \\ \hline f'(x) &=& -\sin(x) \\ &=& \sin(-x) \\ &=& \cos\left(\dfrac{\pi}{2}-(-x)\right) \\ &=& \cos\left(\dfrac{\pi}{2}+x\right) \\ &=& \cos\left(1\cdot \dfrac{\pi}{2}+x\right) \\ \hline f''(x) &=& -\sin\left(1\cdot \dfrac{\pi}{2}+x\right) \\ &=& \sin\left(-(\dfrac{\pi}{2}+x) \right) \\ &=& \cos\left(\dfrac{\pi}{2}-(-(\dfrac{\pi}{2}+x)) \right) \\ &=& \cos\left(\dfrac{\pi}{2}+\dfrac{\pi}{2}+x \right) \\ &=& \cos\left(2\cdot \dfrac{\pi}{2}+x \right) \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline f(x) &=& \cos\left(0\cdot \dfrac{\pi}{2} + x \right) \\ f'(x) &=& \cos\left(1\cdot \dfrac{\pi}{2}+x\right) \\ f''(x) &=& \cos\left(2\cdot \dfrac{\pi}{2}+x \right) \\ \cdots \\ \mathbf{f^n(x)} &=& \mathbf{\cos\left(n\cdot \dfrac{\pi}{2}+x\right)} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{f^n(x)} &=& \mathbf{\cos\left(n\cdot \dfrac{\pi}{2}+x\right)} \\\\ f^{101}(x) &=& \cos\left(101\cdot \dfrac{\pi}{2}+x\right) \quad & | \quad 101\cdot \dfrac{\pi}{2} = 25\cdot (2\pi) + \dfrac{\pi}{2} \\ &=& \cos\left(25\cdot (2\pi) + \dfrac{\pi}{2}+x\right) \\ &=& \cos\left(\dfrac{\pi}{2}+x\right) \\ \mathbf{f^{101}(x)} &=& \mathbf{-\sin\left(x\right) } \\ \hline \end{array}\)
