Question #341951

Use logarithmic differentiation to prove D9: d/dx (1/u^n) =( -n/u^n+1)(du/dx)

Expert's answer

f(x)=1unf(x)=\frac{1}{u^n}

ln⁡f(x)=ln⁡1un=−nln⁡u\ln{f(x)}=\ln{\frac{1}{u^n}}=-n\ln u

ddxln⁡f(x)=ddx(−nln⁡u)\frac{d}{dx}\ln{f(x)}=\frac{d}{dx}(-n\ln u)

1f(x)df(x)dx=−n1ududx\frac{1}{f(x)}\frac{df(x)}{dx}=-n\frac{1}{u}\frac{du}{dx}

df(x)dx=−nf(x)1ududx=−n1un+1dudx\frac{df(x)}{dx}=-nf(x)\frac{1}{u}\frac{du}{dx}=-n\frac{1}{u^{n+1}}\frac{du}{dx}

ddx1un=−n1un+1dudx\frac{d}{dx}\frac{1}{u^n}=-n\frac{1}{u^{n+1}}\frac{du}{dx}



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