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What is the derivative of #x^(lnx)#? - Socratic 6 Nov 2016 · The derivative of x^(lnx) is [(2*y*(lnx)*(x^(lnx)))/x] let y =x^(lnx) There are no rules that we can apply to easily differentiate this equation, so we just have to mess with it until we find an answer. If we take the natural log of both sides, we are changing the equation. We can do this as long as we take into account that this will be a completely new equation: lny=ln(x^(lnx)) …
calculus - Derivative of $\ln |x|$ in the distributional sense ... 31 Mar 2016 · Consider the function $\ln |x|$, since it is locally integrable we can form the distribution $$(\ln |x|,\phi)=\int_{-\infty}^{\infty}\ln |x|\phi(x)dx.$$ Now, I want to show that in the sense of
How do you differentiate f(x)=xlnx-x? - Socratic 14 Nov 2015 · What is the derivative of #f(x)=sqrt(1+ln(x)# ? What is the derivative of #f(x)=(ln(x))^2# ? See all questions in Differentiating Logarithmic Functions with Base e
How do you find the derivative of # (ln(ln(ln(x))) - Socratic 1 Sep 2016 · Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Calculators. 1 Answer . A. S. Adikesavan
What is the derivative of #y=ln(ln(x))#? - Socratic 20 Mar 2018 · (dy)/(dx) = 1/(xlnx) d/dx ln f(x) = ( f'(x) ) / f(x) => d/dx( ln ( ln x ) ) = (d/dx( lnx )) /lnx = (1/x)/lnx 1/( xlnx )
How do you differentiate y = (ln x)^ln x? - Socratic 30 May 2017 · If we do some cancellation we get: #1/x+ln(lnx)/x#, but since they both have denominators of x we can combine them to get #(ln(lnx)+1)/x#. THIS is the derivative of the original exponent which we will multiply with the original function.
How do you find the derivative of ln x? - Socratic 18 Apr 2015 · What is the derivative of #f(x)=sqrt(1+ln(x)# ? What is the derivative of #f(x)=(ln(x))^2# ? See all questions in Differentiating Logarithmic Functions with Base e
How do you find the derivative of #ln(tanx)#? - Socratic 29 Apr 2015 · Use the chain rule and use d/dx(lnu) = 1/u (du)/dx. We'll also need d/dx(tanx) = sec^2x d/dx(ln(tanx))=1/tanx d/dx(tanx) = 1/tanx sec^2x We are finished with the calculus, but we can rewrite the answer using trigonometry and algebra: d/dx(ln(tanx))= 1/(sinx/cosx) 1/(cos^2x)= 1/sinx 1/cosx = cscx secx
Proof of the derivative of $\\ln(x)$ - Mathematics Stack Exchange 28 Jun 2015 · Proof of the derivative of $\ln(x)$ Ask Question Asked 9 years, 8 months ago. Modified 3 years, 4 months ago.
How do you find the derivative of # ln (lnx) - Socratic 30 May 2016 · You have to apply the chain rule that tells us \frac{d}{dx}f[g(x)]=f'[g(x)]g'(x). The f here is the external ln, while the g is the internal ln(x). The derivative of the logarithm is \frac{d}{dx}ln(x)=1/x so the f'[g(x)]=1/ln(x) and the g'(x)=1/x.