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Apply the formula: $\frac{d}{dx}\left(a^b\right)$$=y=a^b$, where $d/dx=\frac{d}{dx}$, $a=x-1$, $b=\cos\left(2x\right)$, $a^b=\left(x-1\right)^{\cos\left(2x\right)}$ and $d/dx?a^b=\frac{d}{dx}\left(\left(x-1\right)^{\cos\left(2x\right)}\right)$
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$y=\left(x-1\right)^{\cos\left(2x\right)}$
Learn how to solve equations rationnelles problems step by step online. d/dx((x-1)^cos(2x)). Apply the formula: \frac{d}{dx}\left(a^b\right)=y=a^b, where d/dx=\frac{d}{dx}, a=x-1, b=\cos\left(2x\right), a^b=\left(x-1\right)^{\cos\left(2x\right)} and d/dx?a^b=\frac{d}{dx}\left(\left(x-1\right)^{\cos\left(2x\right)}\right). Apply the formula: y=a^b\to \ln\left(y\right)=\ln\left(a^b\right), where a=x-1 and b=\cos\left(2x\right). Apply the formula: \ln\left(x^a\right)=a\ln\left(x\right), where a=\cos\left(2x\right) and x=x-1. Apply the formula: \ln\left(y\right)=x\to \frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(x\right), where x=\cos\left(2x\right)\ln\left(x-1\right).
