# Logarithmic Differentiation

• Logarithmic differentiation is a powerful technique to differentiate functions of the form f(x) = [u(x)]v(x). Here both f(x) and u(x) need to be positive for this technique to make sense.

### View the Topic in this video From 32:43 To 49:02

Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers.

• \frac{d}{dx}\left(f(x)^{g(x)}\right)=f(x)^{g(x)}\left[g(x)\frac{f'(x)}{f(x)}+g'(x) \log \ f(x)\right]
• \frac{d}{dx}\left(f(x)^{f(x)}\right)=f(x)^{f(x)}\left[(1+ \log \ f(x))f'(x)\right]
• If y=(f(x))^y \Rightarrow \frac{dy}{dx}=\frac{y^{2}f'(x)}{f(x)[1-y \ \log(f(x))]}
• \frac{d}{dx}\left(\cos^{n}x \ \cos \ nx\right)=n\cos^{n-1}x \ \cos (n+1)x
• \frac{d}{dx}\left(\sin^{n}x \ \sin \ nx\right)=n\sin^{n-1}x \ \sin (n+1)x
• \frac{d}{dx}\left(\cos^{n}x \ \sin \ nx\right)=-n\cos^{n-1}x \ \sin (n+1)x
• \frac{d}{dx}\left(\sin^{n}x \ \cos \ nx\right)=-n\sin^{n-1}x \ \cos (n+1)x
• \frac{d}{dx}\left(Variable^{Variable}\right)=\frac{d}{dx}(Variable)^{Constant}+\frac{d}{dx}(Constant)^{Variable}