
How do you differentiate y = (ln x)^ln x? - Socratic
2017年5月30日 · This means the derivative of #ln(lnx)# is #1/(x*lnx)#. This gives us the derivative of #ln(lnx)*lnx# which is #lnx/(x*lnx)+ln(lnx)/x#. 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 ...
How do you differentiate y=x^lnx? - Socratic
2016年4月18日 · y'=2x^(ln(x)-1)ln(x) Using logarithmic and implicit differentiation, take the natural logarithm of both sides. ln(y)=ln(x^ln(x)) Simplify the right hand side using the rule: ln(a^b)=b*ln(a) ln(y)=ln(x)*ln(x) ln(y)=(ln(x))^2 Take the derivative of both sides. Recall that the chain rule is in effect on both sides.
What is the derivative of #y=ln(ln(x))#? - Socratic
2018年3月20日 · (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 find the inverse of y = ln x? - Socratic
2018年4月4日 · y^-1 = e^x y = ln(x) To find an inverse function, we swap x and y and solve for y again. x = ln(y) e^x = e^(ln(y)) y = e^x Hence: y^-1 = e^x
How do you differentiate y=lnx? - Socratic
2017年1月20日 · 1/x change to exponent form then differentiate as follows: y=lnx=>x=e^y differentiated wrt y (dx)/(dy)=e^y :.(dy)/(dx)=1/e^y=1/x
What is the derivative of y=lnx/ x? - Socratic
2016年1月29日 · Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e. 1 Answer . mason m
How do you find the derivative of y=x^lnx? | Socratic
2016年8月18日 · By taking the natural logarithm of both sides: lny = ln(x^(lnx)) Differentiate both sides: d/dx(lny) =d/dx(lnx(lnx)) 1/y(dy/dx) = square Inset: square We need to differentiate lnx(lnx). By the product rule: [lnx(lnx)]' = 1/x xx lnx + 1/x xx lnx = lnx/x + lnx/x = (2lnx)/x dy/dx =( (2lnx)/x)/(1/y) dy/dx = (2lnx)/x xx y dy/dx = (2lnx)/x xx x^(lnx) Hopefully this helps!
How do you find the derivative of (ln x) ^ cos x? | Socratic
2016年7月13日 · y' = (ln x)^(cos x)[-sin x * ln(ln x) + (cos x)/(x ln x)] For this particular, we'd have to use logarithmic differentiation, which works as follows: Let y = (ln x)^(cos x) Taking the natural log (ln) of both sides yields ln y = ln((ln x)^(cos x)) ln y = cos x * ln(ln x) Since the next step is to take derivatives, the rules we're going to use is d/dx[ln u] = (u')/(u) Differentiating both sides ...
What is the domain and range of #y=ln(x)#? - Socratic
2016年2月8日 · Domain: x>0 Range: -oo<x<oo The graph of y=ln x x can only be all positive and y can take positive or negative values graph{y=ln x[-10, 10, -5,5]} Have a nice day from the Philippines!
How do you find the volume bounded by y=ln(x) and the lines …
2017年8月7日 · For the solution by cylindrical shells, see below. Here is a picture of the region and a representative slice taken parallel to the axis of rotation. The slice is taken at some value of x and has thickness dx. So our functions will need to be functions of x Revolving about the y axis will result in a cylindrical shell. The volume of this representative shell is 2pirh " thickness" The …